Question: If f is a continuous mapping of a metric space $X$ into a metric space $Y$, then $f(\overline{E}) \subset \overline{f(E)}$, for every set $E \subset X$.
Attempt:
For any set $A$, we know, $A \subset \overline{A}$ . Therefore, $f(E) \subset \overline{f(E)} \implies E \subset f^{-1}[\overline{f(E)}]$. Now, $f^{-1}[\overline{f(E)}]$ is closed, (property of any continuous function) hence $\overline{f^{-1}[\overline{f(E)}]}=f^{-1}[\overline{f(E)}]$, giving us, $\overline{E} \subset f^{-1}[\overline{f(E)}]$ [Since $A \subset B \implies \overline{A} \subset \overline{B}$] $\implies f(\overline{E}) \subset \overline{f(E)} $.
Is this correct?
Your proof is fine ! A proof with sequences: let $x_0 \in \overline{E}$. Then there is a sequence $(x_n)$ in $E$ such that $x_n \to x_0.$ Then $f(x_0) = \lim f(x_n) \in \overline{f(E)}.$