$y \in[x]$ $\to (x,y) \in R$ $\to (y,x) \in R$ (since $R$ is symmetric) $\to x \in [y]$.
Therefore $$\tag{1} y\in [x] \to x \in [y].$$
Let $x \in [y]$ $\to (y,x) \in R$ $\to (x,y) \in R$ (sine $R$ is symmetric) $\to y \in [x]$. Therefore $$\tag{2} x \in [y] \to y\in [x].$$
According to (1) and (2),
$x \in [y]$ if and only if $y \in [x]$ it means $[x] = [y]$.
Therefore if $y \in [x]\to [x] = [y]$.
Is this proof is correct? I feel my proof is incomplete. Is there another way to prove it?
We need to check $[x]\subseteq[y]$ and $[x]\supseteq[y]$.
Let $z\in[x]$ be arbitrary. Then by definition, $(x,z)\in R$. We already know $(y,x)\in R$, so by the transitive property, $(y,z)\in R$. Thus, $z\in [y]$. This shows $[x]\subseteq[y]$.
The proof for the reverse inclusion is the same.