As a counterexample, consider any topological space $X$ which is not metacompact.

We recall that a topological space $X$ is *metacompact* each each open cover of $X$ has a point-finite refinement. By Theorem 3.5 in the survey "Covering properties" by D.Burke in the Handbook of Set-Theoretic Topology (1984), a topological space $X$ is metacompact if and only if any directed open cover of $X$ has a point-finite refinement. A cover $\mathcal U$ of $X$ is *directed* if for any sets $U,V\in\mathcal U$ there is a set $W\in\mathcal U$ containing the union $U\cup V$.

By this characterization, each topological space $X$ which is not metacompact admits a directed open cover $\mathcal U$ without point-finite refinement. Since $\mathcal U$ is dierected, each finite subset $F\subset X$ is contained in some set $U\in\mathcal U$. Since $\mathcal U$ has no point-finite refinement, any subcover $\mathcal V$ of $\mathcal U$ is not point-finite, which means that the set $\{V\in\mathcal V:x\in V\}$ is infinite for some $x\in X$.

For an example of a space $X$ which is not metacompact, see Example 4.4 of the survey of Burke. According to this example any Mrowka space $\psi(\mathbb N)$ is not meta-Lindelof and hence not metacompact.