Let $\text{Dom}(f)=E$ be compact and $G_f=\{(x,f(x)): x\in E\}$ the graph of $f$.
$G_f$ is compact iff $f$ is continuous on $E$.
Proof: $\Rightarrow$ Let $f$ is not continuous at some point $x_0\in E$. Then $\exists \{x_n\}$ on $E$ such that $x_n\to x_0$ but $f(x_n)\nrightarrow f(x_0)$. It means that exists some subsequence $\{n_k\}$ such that $|f(x_{n_k})-f(x_0)|\geqslant \varepsilon_0$ for some $\varepsilon_0>0$. We'll define function $g:x\mapsto (x,f(x))$. Then $g(x_{n_k})=(x_{n_k},f(x_{n_k}))$ is sequence in $G_f$ which is compact. Thus this sequence has limit point in $G_f$ namely $(p,f(p))$. Hence $\exists \{k_j\}$ such that $g(x_{n_{k_j}})\to (p,f(p)).$ Hence $p=x_0$ and $f(x_{n_{k_j}})\to f(x_0)$ and we get contradiction. Right?
$\Leftarrow$ Let $\{G_{\alpha}\}$ be an open cover of $G_f$ i.e. $G_f\subset \cup _{\alpha}G_{\alpha}$. Consider "projections" of $G_{\alpha}$ to $E$ we define $F_{\alpha}:=\{x\in E: (x,f(x))\in G_{\alpha}\}.$ It's clear that $E\subset \cup _{\alpha}F_{\alpha}$. Also we define function $g:x\mapsto (x,f(x))$ and $g$ is continouos on $E\times f(E)$ since all his components are continuous. Then $g^{-1}(G_{\alpha})=F_{\alpha}$ is open for any $\alpha$. Since $E$ is compact then $\exists$ $\alpha_1, \cdots, \alpha_n$ such that $E\subset F_{\alpha_1}\cup \cdots\cup F_{\alpha_n}$. Then $g(E)\subset g(F_{\alpha_1})\cup\cdots \cup g(F_{\alpha_n})=G_{\alpha_1}\cup\cdots\cup G_{\alpha_n}.$ But $g(E)=G_f$.
Sorry if my topic is repeated but I solved it myself and canyone check my proof please?
I would be very grateful.
Best Answer
Forwards direction
You said in the comments that your definition of "compact" was "every open cover has a finite subcover", and that you have never encountered sequential compactness in general metric spaces. Therefore, you're going to have to prove that if the space is compact, then it has the property that every sequence has a convergent subsequence. Proofs of this are widely available. The step "this sequence has a limit point" relies on this theorem.
Backwards direction
"$g$ is continuous on $E \times f(E)$" is badly phrased: you mean "$g$ is continuous as a function $E \to E \times f(E)$".
Otherwise, I think it's fine.