Prove it by using that $f^{-1}$ of an open set is open

Question

If $f$ is continuous from a metric space $X$ to a metric space $Y$ prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.

My attempt: I have already proved it in two different ways, one using convergent sequence and another using the fact that if $f$ is continuous then the inverse image of a closed set is closed. But I am not satisfied with these two. I am trying to prove it by using inverse image of open set is open. Can somebody give me a hint?

Answer 1

Assume $f$ is continuous. We show that $x\in \overline{A}$, then $f(x)\in\overline{f(A)}$.Let $A$ be a subset of $X$ Let $N$ be a neighborhood of $f(x)$. Then $f^{-1}(N)$ is an open set of $X$ containing $x$; it must intersect in some point $y$. Then $N$ intersects $f(A)$ in the point $f(y)$, so that $f(x)\in \overline {f(A)}$.