The fundamental result proven by Cantor's diagonal argument (as applied to the real numbers) is that:
For any countable subset $S \subset \mathbb R$ of the real numbers, there is a real number $x \in \mathbb R$ such that $x \notin S$.
From this result, it then obviously follows that $\mathbb R$ itself cannot be countable, since otherwise we could choose $S = \mathbb R$ and derive a contradiction.
(Here, "$S$ is countable" means that there exists a surjection $f$ from the natural numbers $\mathbb N$ to $S$, and hence, that we can enumerate all members of $S$ as $S = \{f(1), f(2),f(3),\dotsc\}$.)
The diagonal argument proves the fundamental result quoted above directly and constructibly, by providing an "algorithm"1 that, given an enumeration $f$ of a countable subset $S = f(\mathbb N) \subset \mathbb R$ of the reals, computes a real number $x \in (\mathbb R \setminus S)$.
As you correctly note, for any given countable subset $S \subset \mathbb R$, it's possible to produce several such numbers $x$ by varying the details of the diagonal algorithm, such as the base used to expand the reals into digit sequences, or the specific digits chosen for $x$, if the base is high enough to allow multiple choices.
This should not really be surprising, since, for any countable set of real numbers $S$, there are infinitely (and, in fact, uncountably) many real numbers that are not in $S$, and could therefore be chosen as $x$. Any given instance of the diagonal algorithm (with the digit choice rules fully specified) only picks out one such $x$ for each enumeration $f$, but that's enough for what it's supposed to do.
(Of course, you can always generate more $x$s by modifying the enumeration $f$ — for example, you can take a previously generated $x$, prepend it to the enumeration, and run the algorithm again to get a new real number $x_2 \ne x$ that is also not in $S$, and so on.)
Ps. As alluded to in the comments, when applying the diagonal argument to the real numbers, there are certain practical reasons to choose a base higher than 2 anyway. The problem is that, in any base $b$, certain real numbers (specifically, those of the form $n / b^k$, for some integers $n$ and $k$) have two possible base-$b$ expansions: one that ends in an infinite string of zeros, and one that ends in an infinite string of $(b-1)$-digits (e.g. 9s, for base 10).
One way to make sure that we don't accidentally generate such an alternate representation of a number that does occur in the given enumeration is to avoid ever choosing the digit $(b-1)$ when constructing the base-$b$ expansion of $x$. In base 2, however, we don't have that freedom: if the $i$-th base-$b$ digit of $f(i)$ is 0, the diagonal algorithm has to choose a non-zero $i$-th digit for $x$, but if $b = 2$, the only possible choice is $1 = (2-1)$. Thus, to make sure that the diagonal algorithm really does what it's supposed to, we really should use base 3 or higher.
That said, it's possible to modify the diagonal algorithm in other ways, to make it work in base 2. For example, we may choose all odd-numbered digits of $x$ to be 0, and choose the even-numbered digits such that digit $2i$ of $x$ is not equal to digit $2i$ of $f(i)$. As long as we also make sure that, for any $f(i)$ that has two base-2 expansions, we always choose the one that ends in all zeros, this is sufficient to guarantee that $x \ne f(i)$ for any $i \in \mathbb N$.
1) It's not actually an algorithm in the strict sense used by computer scientists, since it requires infinitely many steps to exactly compute $x$, but it's close enough.
