So far in the books I've read all proofs involving Alexander Lemma to prove Tychonoff's theorem or in general any compact space they use Zorn Lemma argument. So I was wondering if is there any proof available to show that if $X$ is continuum then the hyperspace $2^X$ is compact by using Alexander Lemma.
$2^X=\{A\subseteq X : A \; \text{is closed and non empty }\}$ .
I assume you are using the Vietoris topology which has a subbase consisting of all sets $\{[U]: U \subseteq X \text{ open }\}$, where $$[U] = \{A \in 2^X: A \cap U \neq \emptyset\}$$ and all sets of the form $\{\langle U \rangle: U \subseteq X \text{ open }\}$ with $$\langle U \rangle = \{A \in 2^X: A \subseteq U\}$$.
This is the subbase we use for Alexander's subbase lemma, and it shows that if $X$ is compact then $2^X$ is compact too:
Let $\{[U_i]: i \in I\} \cup \{\langle V_j \rangle: j \in J\}$ be a cover of $2^X$ by subbasic elements.
Consider $C:= X\setminus \bigcup_{i \in I} U_i$. If $C$ is empty, the sets $U_i$ cover $X$ and so finitely many of them cover $X$, say $U_{i_1}, \ldots, U_{i_n}$. Then if $A \in 2^X$, then $A$ is non-empty and contains some $x$ which must lie in some $U_{i_k}$ for $k \in \{1,\ldots,n\}$ and then $A \in [U_{i_k}]$ and so the sets $\{[U_{i_k}]: k=1, \ldots,n\}$ form a finite subcover of the subbasic cover and we're done.
So we're left with the case that $C \neq \emptyset$ and so $C \in 2^X$ and by definition it misses all $U_i$ so is covered by some $\langle V_{j_0} \rangle$ instead. We then cover $X\setminus V_{j_0}$ by finitely many $U_{i_1}, \ldots U_{i_n}$ and similarly to above show that $$\{[U_{i_k}]: k=1, \ldots,n\} \cup \{\langle V_{j_0} \rangle\}$$ is a finite sucover of the subbasic cover.
Another standard application of the subbase lemma is for ordered spaces $X$ in the order topology to see that they are compact iff every subset of $X$ has a supremum. I do that proof here, e.g. So hyperspaces, ordered spaces and Tychonoff are easily shown compact using Alexander's subbase lemma.