$f,g: X \rightarrow \mathbb{R}$ are continuous. Show $f(A) = g(A) \Rightarrow f(\overline{A}) = g(\overline{A})$ where $A \subseteq X$ 
$f,g: X \rightarrow \mathbb{R}$ are continuous. Show $f(A) = g(A) \Rightarrow f(\overline{A}) = g(\overline{A})$ where $A \subseteq X$

My first intuition is to show one inclusion, and the other inclusion will follow directly.
First we show $f(\overline{A}) \subseteq g(\overline{A})$ 
Pick $f(\overline{a}) \in f(\overline{A})$ 
We know that $\overline{a} \in \overline{A} \Rightarrow \forall \; U \subseteq X$ open: $U \cup A \neq \emptyset$ where $\overline{a} \in U$ 
We also know $f,g$ continuous $\Rightarrow f^{-1}(V) = g^{-1}(V)$ are open in $X$ where $V$ is open in $Y$.
However, I'm not sure if this is all of the information required for the proof. Where else could I look?
 A: Assuming you mean $f(x)=g(x)$ for all $x\in A$: 
let $y\in f(\overline A)$. There is an $x\in \overline A$ such that $f(x)=y.$ Then, $y\in f(\overline A)\subseteq \overline {g(A)}$ so there is a sequence $(x_n)\subseteq A$ such that 
$y_n=g(x_n)=f(x_n)\in g(A)$ such that $y_n\to y$. But this means that $g(x_n)\to y$ and so $|g(x)-f(x)|\le |g(x)-f(x_n)|+|f(x_n)-f(x)|=|g(x)-g(x_n)|+|f(x_n)-f(x)|$ and an easy $\delta-\epsilon$ argument shows that $g(x)=f(x)=y$ so $y\in g(\overline A)$. Symmetry gives the other direction for free. 
It's easier with nets: suppose $(x_{\lambda})$ is a net in $A\subseteq X$ such that $x_{\lambda}\to x\in \overline A$. Then, $|g(x)-f(x)|\le |g(x_{\lambda})-g(x)|+|g(x_{\lambda})-f(x)|=|g(x_{\lambda})-g(x)|+|f(x_{\lambda})-f(x)|$ because $g(x_{\lambda})=f(x_{\lambda})$. To finish, note that each of the last terms in the inequality can be made as small as desired by choosing a suitable neighborhood $X\supseteq U\ni x.$
A: Perhaps the statement should be : $f(a) = g(a)$ for all $a\in A$. As it is stated, the question is not correct. We can have functions $f$, $g\colon [0, \infty)\to \mathbb{R}$ so that
$$f((0, \infty) = f([0, \infty) = (0, \infty)\\
g((0, \infty)= (0,\infty),  \  \ g([0, \infty))= [0, \infty)$$
Take for instance $g(x) = x$, but $f(x)= (\sqrt{x}+1) ( \cos^2 x + \cos^2 (\sqrt{2} x)) $.
(the example of $f$ seems contrived, but there are many other choices).
