Space is $T_1$ iff finite subsets are closed: alternative approach To go about proving the first direction $(\Rightarrow)$, we suppose $X$ is $T_1$. It then suffices to show that singletons are closed and then to rewrite a finite subset as the union of closed singletons. The finite union of finite sets is finite and the finite union of closed sets is closed. Therefore, finite subsets of a $T_1$ space are closed.
My question comes down to this: In order to show $(\Leftarrow)$, is it logically-appropriate to take the $(\Rightarrow)$ proof and just run it backwards?
Let me try it: $(\Leftarrow)$
Suppose for every finite subset $J \subset X$, $J$ is closed.
Since $J$ is finite, $J$ is the finite union of finite sets. Since $J$ is closed, $J$ is the union of closed sets. Combining these two, we can rewrite $J$ as
$$J = \bigcup_{j \in J}\{j\}$$
Since $\{j\}$ is a closed singleton, every singleton subset of $X$ is closed. Thus, $X$ is $T_1. \hspace{0.8cm} \square$

I feel like I'm cheating or doing something wrong here. I literally just ran the proof backwards. And I understand that sometimes you're able to do that with biconditionals but it seems too easy to be true.
Another thing that's wrong with my proof is that I merely showed that $\{j\}$ are closed singletons, but I don't think this generalizes to the entire space $X$ (what if $X$ is infinite?), which is what we need to show.
 A: Sometimes this can work, but more commonly it can't. I don't think your reversal works in this case, though it's a short enough proof that it might look like it works at first glance.
To be clear, I think you're proving the statement

All singleton sets are closed $\iff$ all finite sets are closed.

The $\impliedby$ direction can be done much more quickly than reversing the $\implies$ direction. All singleton sets are finite. If all finite sets are closed, then in particular all singleton sets are.
But when you write the statement

Since $J$ is closed, $J$ is the union of closed sets. Combining these two, we can rewrite $J$ as $$J = \bigcup_{j \in J}\{j\}$$

and conclude from this that $\{j\}$ is closed, you are making a mistake. The union of closed sets is also a closed set; that doesn't mean that if closed set $J$ is the union of some things, those things are closed. (For example: in $\mathbb R$, $[0,1]$ is the union $[0, \frac23) \cup (\frac13, 1]$, but these are not closed sets.)
This seems like the sort of mistake that will happen lots of times you reverse a proof. You can go from a general statement to a specific ("All horses have four legs, therefore this horse has four legs") but when you reverse it, you will be going from a specific statement to a general ("This horse has four legs, therefore all horses have four legs") which isn't valid.
