$f: X \rightarrow Y$ is continuous between two topological spaces. Show $\forall A \subseteq X$ we have $f(\overline{A}) \subseteq \overline{f(A)}$ $f: X \rightarrow Y$ is continuous between two topological spaces. Show $\forall A \subseteq X$ we have $f(\overline{A}) \subseteq \overline{f(A)}$


Since we know that the function is continuous, we know that the preimage of a closed set in $Y$ is a closed set in $X$. 

However, I don't see how this helps.

Another idea I have is to show that if every open set $V \subseteq Y$ containing $x$ where $x \in f(\overline{A})$ we have: $V \cup f(A) \ne \emptyset$.
That would imply that $x \in \overline{f(A)}$

However, I'm not quite sure if this is the most intuitive apprach.

Where would I start to show this inclusion?
You're on the right path but it needs more details in this implication you made (assuming that there is a typo in your question, ie, you wanted to write $f(A) \cap V \neq \emptyset$). What you need to show is: Given $y \in f(\overline{A})$, for every open set $V \subset Y$ such that $y \in V$ you have $V \cap f(A) \neq \emptyset$. This is true since $y \in f(\overline{A})$ implies that $y = f(x)$ with $x \in \overline{A}$, as $f^{-1}(V)$ is an open set (f is a continuous function) and $x \in f^{-1}(V)$ there exists $x_{1} \in f^{-1}(V) \cap A$ which gives you that $f(x_{1}) \in V \cap f(A)$, ie, $V \cap f(A) \neq \emptyset$.
Let $f(B)=\overline{f(A)}$, then by continuity B is closed and $A\subseteq B$. As $\overline{A}$ is an intersection of all closed sets containing $A$, we get $\overline{A}\subseteq B$ and $f(\overline{A})\subseteq f(B)=\overline{f(A)}$