Solution: If the topology $\mathcal{T} $ on $X$ is Hausdorff, then this is not possible. For proving this, let $K$ be a compact subset of $X$. We claim that $\bar{K} $ is compact. Note that $\bar{K} $ is closed and the $\mathcal{T} $ is Hausdorff, closed subset of a compact set is compact. Hence, $\bar{K} $ is compact.

Now let us construct an example. Let $\star$ be any point. Take $X = \{p\} \cup \mathbb{N}$ and define the topology $\mathcal{T} $ on $X$ as follows: \[ \mathcal{T} = \left\{ U : p \in U \right\} \cup \{ \emptyset \}. \] It is clear that $\mathcal{T} $ is a topology on $X$ and $\overline{\{ p \} } $ is a compact subset of $X$. Also, note that $ \overline{\{ p \} } = X$, which is not compact. Now it remains to show that $X$ is not compact. Consider the open cover of $X$ defined as $\mathcal{U} = \{ U_n : n\in \mathbb{N} \}$, where $U_n = \{ p,n \}$. The open cover $\mathcal{U} $ does not have a finite subcover. Thus, $X$ is not compact.