The Sorgenfrey plane is not metacompact by [Example 2 in these notes][1]. However, the Sorgenfrey plane is the product of two paracompact (and hence metacompact) spaces. Therefore metacompact spaces are not even closed under taking finite products.

By transfinite induction, one can show that every open cover of the Sorgenfrey line $S$ has a refinement that partitions $S$ into intervals of the form $[a,b)$.

Let me now give an explicit example with proof of a cover of the Sorgenfrey plane $S\times S$ with no point finite open refinement. 

For each real number $x$, let $U_{r}=[r,\infty)\times[-r,\infty)$ and let $U=\{(x,y)\in S\times S|x+y<0\}$. Let $\mathcal{U}=\{U\}\cup\{U_{r}|r\in S\}$. Then $\mathcal{U}$ is an open cover of $S\times S$. I however claim that $\mathcal{U}$ has no point-finite open refinement. Suppose that $\mathcal{V}$ is an open refinement of $\mathcal{U}$. Then for each $r\in S$, there is some $V_{r}\in\mathcal{V}$ with $(r,-r)\in V_{r}$ and each $V_{r}$ is distinct. Therefore, for each $r\in S$, there is some $\epsilon_{r}>0$ such that $[r,r+\epsilon_{r})\times[-r,-r+\epsilon_{r})\subseteq V_{r}$. Let $A_{n}=\{r\in\mathbb{R}|\epsilon_{r}<\frac{1}{n}\}$. Then there is some $A_{n}$ which is non-meager. In particular, $A_{n}$ is not nowhere dense, so there is some interval $I$ so that $A_{n}\cap I$ is a dense subspace of $I$. However, since $A_{n}\cap I$ is a dense subspace of $I$, if $r\in I$, then we have $\langle r+\frac{1}{2n},-r+\frac{1}{2n}\rangle\in [s,s+\epsilon_{s})\times[-s,-s+\epsilon_{s})\subseteq V_{s}$ whenever $s\in I\cap A_{n}\cap(r-\frac{1}{2n},r+\frac{1}{2n})$. Since $\langle r+\frac{1}{2n},-r+\frac{1}{2n}\rangle$ is contained in infinitely many sets of the form $V_{s}$, the refinement $\mathcal{V}$ is not point finite. Therefore $X$ is not metacompact.





  [1]: https://dantopology.wordpress.com/tag/metacompact-space/