Proof Attempt:
Consider the open cover $\{N_{\epsilon_i}(x_i)\}_{x_i\in X}$ where $N_{\epsilon_i}(x_i)$ is the open $\epsilon$-neighborhood of $x_i \in X$. Assuming that $X$ is compact, $\{N_{\epsilon_i}(x_i)\}_{x_i\in X}$ would then have a finite subcollection, $\{N_{\epsilon_i}(x_i)\}_{i=1}^n$, such that $X \subseteq \bigcup_{i=1}^n N_{\epsilon_i}(x_i)$. So if $(x_k)$ is any sequence in $X$, then for all $k \in \Bbb{N}$, there exists $i=1,2,\cdots,n$ where $x_k\in N_{\epsilon_i}(x_i)$. Since $\epsilon$ is an arbitrary positive real number, then as $\epsilon \to 0$, the preceding argument would still hold; that is $\forall\epsilon\gt 0, \forall k\in\Bbb{N}, \exists i=1,\cdots n$, such that $x_k\in N_{\epsilon_i}(x_i)$. Therefore, $(x_n)$ has at least one cluster point.
Best Answer
No, it's not really correct and the notation you're using is very confusing.
Given a sequence $(z_k)_{k\in\mathbb{N}}$ you need to find $x$ such that, for every $\varepsilon>0$, infinitely many terms of the sequence belong to $N_\varepsilon(x)$.
Assume the contrary: for every $x\in X$, there is $\varepsilon_x>0$ such that $N_{\varepsilon_x}(x)$ contains only finitely many terms of the sequence: precisely, there exists $k_x$ such that $z_k\notin N_{\varepsilon_x}(x)$, for every $k>k_x$.
The family $(N_{\varepsilon_x}(x))_{x\in X}$ is an open cover of $X$, so it admits a finite subcover, say $$ X=\bigcup_{i=1}^n N_{\varepsilon_i}(x_i) $$ (where $\varepsilon_i=\varepsilon_{x_i}$ for readability).
Consider $\bar{k}=\max\{k_{x_1},k_{x_2},\dots,k_{x_n}\}$. For $k>\bar{k}$, we have by construction that $z_k\notin N_{\varepsilon_i}(x_i)$, which is a contradiction.
A comment. Such compactness arguments are highly non constructive, so you cannot hope to find a particular cluster point. At most you can prove, as done before, that assuming no cluster point exists leads to a contradiction.