Why is it that $x\sim y$ implies that $[x]=[y]$?

Question

I had a doubt in the proof of this property:

For $x\in A$, let $[x]$ be the set $[x]=\{a\in A\mid a\sim x\}$.

a) $[x]=[y]$ or

b) $[x]\cap[y]=\varnothing$

I understand till the part where they prove that if for some $c$, $c\sim x$ and $c\sim y$ then $x\sim y$. But then the conclusion that $x\sim y$ implies $[x]=[y]$ strikes a bit of confusion.

I need help regarding this. Thank you very much.

Answer 1

If you understood that far, you're practically done. If $x\sim y$, then take $c\in[x]$, then $c\sim x$ and therefore $c\sim y$, since $x\sim y$, so $c\in[y]$. Therefore $[x]\subseteq[y]$. The argument for $[y]\subseteq[x]$ is similar, and therefore $[x]=[y]$.