Proofs I saw used the Lebesgue number or Lindelof property. I used total-boundedness and I wonder if this is correct.
Proof attempt: By a known result. If $X$ is sequentially compact then it is totally bounded so $\forall \epsilon \gt 0, \exists A=\{x_1,\cdots,x_k\}\subseteq X$ such that $\forall x \in X, \exists i=1,\cdots,k$ where $\operatorname{d}(x,x_i) \lt \epsilon$ which equivalently means $x \in \operatorname{B}_{\epsilon}(x_i)$. This implies $X \subseteq \bigcup_{i=1}^{k}\operatorname{B}_\epsilon (x_i)$. If $\{O_j\}_{j \in I}$ is an open cover of $X$, then for all $x \in X$ there exists $j \in I$ such that $x \in O_j$ where for some $\epsilon' \gt 0$, $\operatorname{B}_{\epsilon'}(x) \subseteq O_j$. Since $X$ is totally bounded then for some $n \in \Bbb{N}$ we have $X \subseteq \bigcup_{i=1}^{n}\operatorname{B}_{\epsilon'}(x_i)$ with respect to $\epsilon' \gt 0$ where for all $i=1,\cdots,n$ there exists $j \in I$ such that $\operatorname{B}_{\epsilon'}(x_i) \subseteq O_j$. By picking all elements of $\{O_j\}_{j \in I}$, where an open ball $\operatorname{B}_{\epsilon'}(x_i)$ is contained, to be in a set called $\Omega$, then $\Omega$ is the finite subcover of $\{O_j\}_{j \in I}$.
No, it is not correct. You wrote that “for all $x \in X$ there exists $j \in I$ such that $x \in O_j$ where for some $\varepsilon' \gt 0$, $B_{\varepsilon'}(x) \subseteq O_j$”. Yes, but $\varepsilon'$ is not a fixed number greater than $0$; it depends on $x$. Therefore, you cannot deduce that “we have $X \subseteq \bigcup_{i=1}^{n}B_{\varepsilon'}(x_i)$”.