I think out one proof that integers are not countable, while integers are actually countable, so I don't know where my proof is wrong:
Suppose the set of integers is $A$. Choose one countable subset of integers, name it $s$, and the elements of $s$ are $s_1,s_2,...$The elements in $s$ are different series, which are constructed with numbers from 0 to 9. Now we can find another integer $t$ out of the set $s$ in such a way: the ith element of the $t$ is different from the ith element of $s_i$. Hence after the construction, $t$ is different from any element of $s$, and $s$ is a proper subset of $A$. Since every countable subset of $A$ is a proper subset of $A$, then $A$ can't be a countable set itself.
Cantor's proof of the uncountability of the reals is what I'm assuming you want to mimic. Here's how it goes. Even though you probably already know this, it's helpful to point out the parallel.
Suppose you could enumerate $\mathbb{R}$. So, list them out as $r_1,r_2,r_3,r_4,...$ and write out their decimal expansion. I'm going to write out some explicit forms of these with specific digits to illustrate the point.
$r_1 = 9.14234092357983454424954289432...$
$r_2 = 8.923749823749328579328586852435...$
$r_3 = 0.298723987589327492342357235963...$
$r_4 = 1.2384798324732834657832576428546...$
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We can construct a new real number that is not in this list, by changing the digit of $r_i$ in the $i^{th}$ spot. This number is not in our list, and thus we cannot enumerate $\mathbb{R}$.
Here's how your proof goes. Enumerate the integers by their digits, like you did the reals. Call the integers $x_1,x_2,x_3,...$
$x_1 = 232987298571983751982357192835...$
$x_2 = 691578941384757893728945974823...$
$x_3 = 9845479843758239874719382175097...$
THIS is where your proof already fails. This isn't a list of integers. Each of these numbers has INFINITELY many nonzero digits. Integers have a finite number of digits.