I am currently brushing up on some very basic set theory, as I've been away from proof writing and such for a few years. The statement I wrote a proof for is:
If $\sim$ is an equivalence relation on a set $X$ then $[x]\neq[y]\iff[x]\cap[y]=\emptyset$.
For context, I wrote this statement myself (another skillset I am rebuilding) and recognize that it comes off as a bit awkward (although I believe it is a true statement). In any case, here is the proof I wrote. I am looking for general advice regarding clarity and conciseness, as well as any suggestions on what I may add or remove to make this proof stronger. Of course, if there are any errors please point those out!
Proof.
($\Rightarrow$) Let $X$ be a set with equivalence relation $\sim$. For $x,y\in X$ assume that $$[x]\neq[y].$$ Suppose towards a contradiction that $[x]\cap[y]\neq\emptyset$. If this intersection is nonempty, then there is some $z$ such that $z\in[x]$ and $z\in[y]$. By definition of equivalence class, $z\in[x]$ implies that $z\sim x$ and, similarly, $z\in[y]$ implies that $z\sim y$. Since $\sim$ is transitive, this means that $x\sim y$. Therefore $x\sim a$ for all $a\in[y]$ and $y\sim b$ for all $b\in[x]$. Thus $[x]=[y]$, a contradiction.
($\Leftarrow$) Suppose that $[x]\cap[y]=\emptyset$. Assume towards a contradiction that $[x]=[y]$.
If $z\in[x]$ then $z\in[y]$ by definition of set equality. Thus $z\in[x]\cap[y]$, a contradiction. $\blacksquare$
Thank you for your time and assistance.
Overall this is a good solution. As mentioned in the other answer, some more details on how $z\sim x$ and $z\sim y$ implies $x\sim y$, in particular through symmetry and transitivity, would be a good idea. Go through it one step at a time, making note of exactly which property you're using where. And also how this implies that $x\sim a$ for all $a\in [y]$ and $y\sim b$ for all $b\in [x]$.
As a matter of aesthetics, I personally don't like to use contradiction where it is unnecessary. In your first part, you only really use $[x]\cap [y]\neq\emptyset$, and conclude that $[x]=[y]$. No need for contradiction there, no need to assume $[x]\neq [y]$. What you have is just a pure proof by contraposition. Similarly for the second part you just need to assume the two classes are equal, and you can show that the intersection is non-empty (as it clearly contains, at the very least, $x$). No need to also assume that the intersection is empty.
Which is to say, you have actually proven $$[x]=[y]\iff [x]\cap [y]\neq\emptyset$$by direct proof.