This is an overkill, but you can show something stronger: it is possible to go between any two vertices in an uniformly bounded number of steps, i.e. there is a constant $K$ such that any two vertices are connected by a path of length at most $K$.
A natural $a$ is called a Polignac number if there are infinitely many pairs of primes whose difference is $a$.
Assume $y>x$. Note that you can go from $x$ to $y$ in two steps if there's $z$ such that $x+z,y+z$ are prime. As an important case, you can go from $x$ to $y$ if $y-x$ is a Polignac number.
Pintz has recently shown (page 3) that there's a constant $C$ such that every interval $[n,n+C]$ contains a Polignac number.
To prove that there's a path from $x$ to $y$, take a Polignac number $x-a \in [x-C,x]$ and go from $x$ to $a$ (note their difference is a Polignac number and $a \leq C$). Likewise, go from $y$ to $b$ (where $b \leq C$).
Now you need to connect $a$ to $b$, where both numbers are bounded by $C$. You can do this in a bounded number of steps via the Bertrand's postulate.