Continuous image of the closure of a set in a compact space is closure of image
Solution 1:
The set $\overline{A}$ is compact as the closed subset of a compact space. Continuous functions map compact sets to compact sets. So $\pi(\overline{A})$ is compact. At least if $Y$ is a Hausdorff space, $\pi(\overline{A})$ is then also closed. So $\pi(\overline{A})$ is a closed set containing $\pi(A)$. Since $\overline{\pi(A)}$ is the smallest closed set containing $\pi(A)$, we have $\overline{\pi(A)}\subseteq \pi(\overline{A})$.