Have guest $i$ on day $j$ stay in room number $p_i^j$, where $p_i$ is the $i$'th prime. The fundamental theorem of arithmetic shows that each guest stays in a unique room compared to all other guests and to their own previous occupied rooms.

Update: As reminded by Will Orrick's comment below, I should have been clear about how to appropriately assign specific guest identifiers ($i$ used above) for the "infinitely many coaches with infinitely many members in each coach". This assumes all of the coaches arrive & can be accessed at the same time, as even the tiniest finite delay in time between coaches means not all of the coaches can arrive in one day. With this assumption, let $c_{m,n}$ be the $n$'th member of coach $m$, with $m,n \ge 1$. Next, order the coach members as shown below

$$\begin{array} 0c_{1,1} & c_{1,2} & c_{1,3} & c_{1,4} & c_{1,5} & \dots\\ c_{2,1} & c_{2,2} & c_{2,3} & c_{2,4} & c_{2,5} & \dots\\ c_{3,1} & c_{3,2} & c_{3,3} & c_{3,4} & c_{3,5} & \dots\\ c_{4,1} & c_{4,2} & c_{4,3} & c_{4,4} & c_{4,5} & \dots\\ c_{5,1} & c_{5,2} & c_{5,3} & c_{5,4} & c_{5,5} & \dots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$$

Assign each guest number in turn, start at $i = 1$ at $c_{1,1}$, then going right to $c_{1,2}$ (for $i = 2$), down to $c_{2,2}$ (for $i = 3$), left to $c_{2,1}$, then down to $c_{3,1}$, right until $c_{3,3}$, then up until $c_{1,3}$, then right to $c_{1,4}$, then down to $c_{4,4}$, then left to $c_{4,1}$, etc. As you can see, this snakelike type pattern will cover all of the entries above, and will eventually get to any $c_{m,n}$ coach member exactly once. This shows it's a bijection between the assigned guest numbers and all of the infinitely many coach members among all of the infinitely many coaches.

Note the method I used is similar to some arguments used to prove the cardinality of the set of rational numbers is the same as the set of the integers, like what this answer of All sets of rational numbers are bigger than the set containing infinite integers - or are they? uses.

Have guest $i$ on day $j$ stay in room number $p_i^j$, where $p_i$ is the $i$'th prime. The fundamental theorem of arithmetic shows that each guest stays in a unique room compared to all other guests and to their own previous occupied rooms.

Update: As reminded by Will Orrick's comment below, I should have been clear about how to appropriately assign specific guest identifiers ($i$ used above) for the "infinitely many coaches with infinitely many members in each coach". This assumes all of the coaches arrive & can be accessed at the same time, as even the tiniest finite delay in time between coaches means not all of the coaches can arrive in one day. With this assumption, let $c_{m,n}$ be the $n$'th member of coach $m$, with $m,n \ge 1$. Next, order the coach members as shown below

$$\begin{array} 0c_{1,1} & c_{1,2} & c_{1,3} & c_{1,4} & c_{1,5} & \dots\\ c_{2,1} & c_{2,2} & c_{2,3} & c_{2,4} & c_{2,5} & \dots\\ c_{3,1} & c_{3,2} & c_{3,3} & c_{3,4} & c_{3,5} & \dots\\ c_{4,1} & c_{4,2} & c_{4,3} & c_{4,4} & c_{4,5} & \dots\\ c_{5,1} & c_{5,2} & c_{5,3} & c_{5,4} & c_{5,5} & \dots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$$

Assign each guest number in turn, start at $i = 1$ at $c_{1,1}$, then going right to $c_{1,2}$ (for $i = 2$), down to $c_{2,2}$ (for $i = 3$), left to $c_{2,1}$, then down to $c_{3,1}$, right until $c_{3,3}$, then up until $c_{1,3}$, then right to $c_{1,4}$, then down to $c_{4,4}$, then left to $c_{4,1}$, etc. As you can see, this snakelike type pattern will cover all of the entries above (by going through all entries of the edges of an increasing square-like pattern), and will eventually get to any $c_{m,n}$ coach member exactly once. This shows one bijection (among other possible ones) between the assigned guest numbers and all of the infinitely many coach members among all of the infinitely many coaches.

Note the method I used is similar to some arguments used to prove the cardinality of the set of rational numbers is the same as the set of the integers, like what this answer of All sets of rational numbers are bigger than the set containing infinite integers - or are they? uses.

Have guest $i$ on day $j$ stay in room number $p_i^j$, where $p_i$ is the $i$'th prime. The fundamental theorem of arithmetic shows that each guest stays in a unique room compared to all other guests and to their own previous occupied rooms.

Have guest $i$ on day $j$ stay in room number $p_i^j$, where $p_i$ is the $i$'th prime. The fundamental theorem of arithmetic shows that each guest stays in a unique room compared to all other guests and to their own previous occupied rooms.

Update: As reminded by Will Orrick's comment below, I should have been clear about how to appropriately assign specific guest identifiers ($i$ used above) for the "infinitely many coaches with infinitely many members in each coach". This assumes all of the coaches arrive & can be accessed at the same time, as even the tiniest finite delay in time between coaches means not all of the coaches can arrive in one day. With this assumption, let $c_{m,n}$ be the $n$'th member of coach $m$, with $m,n \ge 1$. Next, order the coach members as shown below

$$\begin{array} 0c_{1,1} & c_{1,2} & c_{1,3} & c_{1,4} & c_{1,5} & \dots\\ c_{2,1} & c_{2,2} & c_{2,3} & c_{2,4} & c_{2,5} & \dots\\ c_{3,1} & c_{3,2} & c_{3,3} & c_{3,4} & c_{3,5} & \dots\\ c_{4,1} & c_{4,2} & c_{4,3} & c_{4,4} & c_{4,5} & \dots\\ c_{5,1} & c_{5,2} & c_{5,3} & c_{5,4} & c_{5,5} & \dots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$$

Assign each guest number in turn, start at $i = 1$ at $c_{1,1}$, then going right to $c_{1,2}$ (for $i = 2$), down to $c_{2,2}$ (for $i = 3$), left to $c_{2,1}$, then down to $c_{3,1}$, right until $c_{3,3}$, then up until $c_{1,3}$, then right to $c_{1,4}$, then down to $c_{4,4}$, then left to $c_{4,1}$, etc. As you can see, this snakelike type pattern will cover all of the entries above, and will eventually get to any $c_{m,n}$ coach member exactly once. This shows it's a bijection between the assigned guest numbers and all of the infinitely many coach members among all of the infinitely many coaches.

Note the method I used is similar to some arguments used to prove the cardinality of the set of rational numbers is the same as the set of the integers, like what this answer of All sets of rational numbers are bigger than the set containing infinite integers - or are they? uses.