$\DeclareMathOperator{\int}{int}$Let $(X,(e_\alpha)_{\alpha \in I})$ be a finite cell complex, i.e. $I$ finite. If $X$ is Hausdorff then the closures of the cells $\bar e_\alpha$ are compact hence $X$ is compact as a finite union of compact spaces.
What happens if $X$ is not Hausdorff? Are the closures of the cells still compact?
If $X$ is not Hausdorff there could be dense $0$-cells. But all finite examples I found were still compact. So is there an example of a finite and non-compact cell complex?
Definition of cell complex:
A cell complex is a pair $(X,(e_\alpha)_{\alpha \in I})$ where $X$ is a topological space and $(e_\alpha)_{\alpha \in I}$ is a cell decomposition, i.e. $e_\alpha\subset X$ and $X = \bigcup_{\alpha \in I} e_\alpha$ disjoint union and there is a map $n: I \to \mathbb{N_0}$ and characteristic maps $\Phi_\alpha: D^{n(\alpha)} \to X$ such that $\Phi_\alpha|_{\int{D}^{n(\alpha)}}$ is a topological embedding with $e_\alpha = \Phi_\alpha(\int{D^{n(\alpha)}})$ and $\Phi_\alpha(\partial D^{n(\alpha)})\subseteq \bigcup_{n(\beta) < n(\alpha)} e_\beta$ for all $\alpha \in I$.
