$f$ is open iff $f(Int(A)) \subset Int(f(A))$

Question

Prove that $f:X \rightarrow Y$ is open iff $f(Int(A)) \subset Int(f(A)), \ A \subset X.$

Right direction:

If $f$ is open then it transforms open sets into open sets. Let $A \subset X$. $Int(A)$ is open in $X$, so $f(Int(A))$ is open in $Y$, which means $f(Int(A))=Int(f(Int(A))) \subset Int(f(A))$.

I can't get to prove the other direction.

Answer 1

Suppose that $f(Int(A))\subset int(f(A))$. Let $U$ be an open subset, $int(U)=U$, we have $f(U)=f(int(U))\subset int(f(U))$, since $int(f(U))\subset f(U)$, we deduce that $f(U)=int(f(U))$ and henceforth $f(U)$ is open.

Answer 2

Your hypotheses says that for any open set $U\subseteq X$, and $x\in U$ that $f(U)$ is a neighborhood of $f(x)$. Therefore $f(U)$ is open.