I'm trying to prove this:
Let $R$ be an equivalence relation in $X$. Show that $A$ is open in $X/R .\iff \bigcup_{[x]\in A}[x]$ is open in $X$
One of the first things that come to my mind is that there may be a typo, the last set shouldn't be $\bigcup_{[x]\in A}x$?.
Still I don't know how to solve this problem. I consider that there is a projection $p:X\to X/R$, then it seems I could take the set and the preimage of it to check if its open.
Trying to prove it I wrote this:
$(1)$ Suppose $A$ is open in $X/R$.
Since $X/R$ has the quotient topology, for $A\subset X/R$ must be $p^{-1}(A)\subset X$ open. Doesn't look like the direct implication follows immediatly?, $A$ is a set of equivalence classes, the union of all equivalence classes inside $A$ wouldn't be $A$ itself?, then $\bigcup_{[x]\in A}x$ must be open in x
$(2)$ Suppose $\bigcup_{[x]\in A}x$ is open in $X$.
This may not look as straight forward as $(1)$ (if it was), but isn't $p$ a projection, then $p$ continous?. If that's the case $p^{-1}$ would carry open sets to open sets and closed sets to closed sets, then by $(1)$ $A$ must be open in $X/R$ otherwise the set $\bigcup_{[x]\in A}x$ would be closed.
There is no typo in the statement. Note that $[x]\subseteq X$ and $x\in X$. A union makes sense only for the former objects in this case.
The definition of the quotient topology says that $U\subseteq X/R$ is open iff $p^{-1}U\subseteq X$ is open. So in your case, $A\subseteq X/R$ is a collection of equivalence classes, and $p^{-1}A=\bigcup_{[x]\in A} [x]$. Why? If $x\in p^{-1}A$ then $[x]=p(x)\in A$, so $x\in [x]\in \bigcup_{[x]\in A}[x]$. Hence $p^{-1}A\subseteq \bigcup_{[x]\in A} [x]$. Can you show the other inclusion?