p closed continuous function $p(\bar A) =\overline {p(A)}$

Question

Is it correct that when $p:X\rightarrow Y $ is a continuous and a closed map then if $A \subset X$: $$p(\bar A)= \overline{p(A)}$$

My attempt: As $p$ is a continuous function, then $p(\bar A) \subseteq \overline{p(A)}$. We have $p$ is closed map so $p(\bar A)$ is a closed subset in $Y$ that $p(A) \subseteq p(\bar A) \subseteq \overline{p(A)}$. We know $\overline{p(A)}$ is the smallest closed in $Y$ that contains $p(A)$. As the $p(\bar A) \subseteq \overline {p(A)}$ then: $$p(\bar A) =\overline {p(A)}$$

Answer 1

Your approach is correct, but I believe it is somewhat misleading to write

We have $p$ is closed map so $p(\bar A)$ is a closed subset in $Y$ that $p(A) \subseteq p(\bar A) \subseteq \overline{p(A)}$. We know $\overline{p(A)}$ is the smallest closed in $Y$ that contains $p(A)$. As ...

In fact, we know that $p(\bar A)$ is a closed subset in $Y$ such that $p(A) \subseteq p(\bar A)$. [The second part $p(\bar A) \subseteq \overline{p(A)}$ is also true because $p$ is continuous, but it is irrelevant here.] Moreover $\overline{p(A)}$ is the smallest closed set in $Y$ that contains $p(A)$, thus $\overline{p(A)} \subseteq p(\bar A)$. [You do not state this fact explicitly.] This allows to conclude $\overline{p(A)} = p(\bar A)$.