Is there a countable space that is $T\_1$ and not metacompact? (A space $(X,\tau)$ is not metacompact iff there is on open cover $\cal{U}\_0$ such that for every open refinement $\cal V$ there is $x\in X$ such that $x$ is contained in infinitely many members of $\cal V$.)


Note that $(\omega,\tau)$ with $\tau=\{\emptyset,\omega\}\cup\big\{\{0,n\}:n\in\omega\big\}$ is $T\_0$ and not metacompact.



Countable, $T_1$, and not metacompact
No, every countable T$\_1$ space is metacompact. Let $\tau$ be any T$\_1$ topology on $\mathbb N.$ Let $\mathcal U$ be any open cover. For each $n\in\mathbb N$ choose $U\_n\in\mathcal U$ with $n\in U$ and let $V\_n=U\_n\setminus\{1,\dots,n-1\}.$ Then $\mathcal V=\{V\_n:n\in\mathbb N\}$ is a point-finite open refinement of $\mathcal U.$


