I will explain how I solved the problem and will appreciate it if you check if my solution is right.
(if $f$ is continuous on $E$, its graph is compact)
Let $\{A_\alpha\}$ be an open cover of the graph.
For each $x \in E$, $\exists \alpha_x$ such that $(x, f(x)) \in A_{\alpha_x}$.
Since $A_\alpha$ is open, $\exists \epsilon_x$ such that an open ball $B((x, f(x)), \epsilon_x) \subset A_{\alpha_x}$
Because $f$ is continuous at x, for the given $\epsilon_x$, $\exists \delta_x>0$ such that if $d(y,x)<\delta_x$, $d(f(x), f(y))<\epsilon_x$. Let $r_x = min(\delta_x, \epsilon_x)$. Then, $\forall y \in B(x, r_x) $, $f(y)\in B(f(x),r_x)$; therefore, $(y, f(y)) \in B((x, f(x)), r_x) \subset B((x, f(x)), \epsilon_x)$. Thus, $\{B(x, r_x)\}$ is an open cover of $E$.
Because $E$ is compact, we can pick $x_1, … x_n$ such that $U^n_{k=1}B(x_k, r_k)=E$. Therefore, the graph $\subset U^n_{k=1}B((x_k, f(x_k)), \epsilon_x) \subset U^n_{k=1} A_{\alpha_{x_k}} $.
(if the graph of $f$ is compact, $f$ is continuous on $E$)
Pick $x \in E$ and an open cover of the graph $\{B_\alpha((x, f(x)), \epsilon_x) \}$. Because the graph is compact, the graph $\subset U^n_{\alpha = 1} B_\alpha((x, f(x)), \epsilon_x)$. If $(y, f(y)) \in B((x, f(x)), \epsilon_x)$, $y \in B(x, \epsilon_x)$ and $f(y) \in B(f(x), \epsilon_x)$.
Choose an open cover of $E$, and because $E$ is compact, $U^m_{\beta=1} B_\beta (x, \delta_x)$. Pick $r_x = min(\epsilon_x, \delta_x)$. If $y \in B(x, r_x), f(y) \in B(f(x), \epsilon_x)$. Thus, $f$ is continuous.
Thank you in advance!
Best Answer
The proof of "$f$ continuous then $\Gamma(f)$ (the graph of $f$) compact" has a right idea, but could be much more easily proved, if you use the fact that the continuous image of a compact space is compact, a thing you're reproving in part for this special case, it seems to me.
The reverse direction is not correctly shown: for your approach you should start with a given $\epsilon>0$ for $x$ and $f(x)$, and find a correct $\delta$ for that $\epsilon$. But easier is to note that $f^{-1}[C] = \pi_1[\Gamma(f) \cap (X \times C)]$ for any closed $C$ in the codomain (and where $\pi_1$ is the first projection onto $E$).