Prove that $\overline{f(A)}\subseteq f(\overline{A})$ where $f: X \rightarrow Y$ is continuous, $X$ is compact and $A \subseteq X$

The claim can be made false without the assumption that $Y$ is Hausdorff. To see this, let $Y\equiv\{1,2\}$ be endowed with the indiscrete topology (that is, the only open sets are the empty set and the whole set). Let $X\equiv\{1,2\}$ be endowed with the discrete topology (all subsets are open) and $f(1)\equiv 1$ and $f(2)\equiv 2$. Clearly, $f$ is continuous. But if $A\equiv\{1\}$ (a finite set is compact in any topological space), then $$\overline{f(A)}=\overline{\{1\}}=\{1,2\},$$ yet $$f(\overline{A})=f(\{1\})=\{1\}.$$

This is false in general if $Y $ is not Hausdorff.

Take any map $f : X \to X $, where on the left, $X$ has the discrete topology and on the right the indiscreet topology (only $\emptyset $ and $X$ are open), where $X $ is any finite set.

Finally, let $A=\{x\} $. Is is not hard to see that this is a counterexample (if $X $ has more than one element).