Let $X$ be a $T_1$ space and let $A \subseteq X$ be any subset.
Fake Proof: Since $X$ is $T_1$ we know that any singleton in $X$ is closed. So choose $a \in A$, then $\{a\}$ is closed and $X \setminus \{a\}$ is open in $X$. Now choose another $a' \in A$ such that $a' \neq a$. Then again $\{a'\}$ is closed and $(X \setminus \{a\}) \setminus \{a'\} = X \setminus \{a, a'\}$ is open since it is the set difference of an open and closed set.
Continue this process for all $a \in A$ and thus we've shown $X \setminus A$ is open and hence that $A$ is closed. $\square$
My guess is that when we say "continue this process" that's where some subtlety lies which would show that this proof is incorrect, but I can't say what that subtlety is. If $A$ was finite this proof would work fine I think.
So my question is, what exactly is wrong with this proof?
You're right. The "continue this process" hides that we get $A$ is closed via $$X\backslash A=\bigcap_{a\in A}(X\backslash\{a\}).$$
If $A$ is infinite, this is an infinite intersection of open sets, which is not necessarily open.