${\bf EXERCISE:}$ Let $X$ be a topological space and $A \subset X$. Suppose that for each $x \in A$, there is some open set $U$ such that $x \in U \subset A$. Prove that $A$ is ${\bf open} $ in $X$.
Proof.
I think this is very easy but maybe I am interprenting the wrong way :
Since $\forall x \in A \exists U $ with$x\in U$ then $A = \bigcup_{x \in U} U $
since $U$ is open in $X$, then the union is also open in $X$ so $A$ is open in $X$. QED
Is this correct?