The answer is yes if $X$ is locally compact. Indeed, consider the topological space $X \times \mathcal{K}$ and let $V = \{(x,U) \in X \times \mathcal{K} | x \in U\}$. We claim that $V$ is open in $X \times \mathcal{K}$. To see this, take $(x,U) \in V$ and let $W$ be an open neighborhood of $x$ such that $\overline{W}$ is compact and contained in $U$. Then the set $W \times (\mathcal{K} \cap \mathcal{V}_{\overline{W}}) \subseteq X \times \mathcal{K}$ is an open neighborhood of $(x,U)$ which is contained in $V$, and so $V$ is open. Let us now show that $\cap_{U \in \mathcal{K}} U \subseteq X$ is open. First observe that $y \in \cap_{U \in \mathcal{K}} U$ if and only if $y \times \mathcal{K}$ is contained in $V$. Since $V$ is open and $\mathcal{K}$ is compact, if $y \times \mathcal{K} \subseteq V$ then there exists an open neighborhood $U_y$ of $y$ such that $U_y \times \mathcal{K} \subseteq V$ (this is a standard compactness argument). We may finally conclude that $U_y \subseteq \cap_{U \in \mathcal{K}} U$ and so $\cap_{U \in \mathcal{K}} U$ is open.

If $X$ is not locally compact then I'm not sure what the answer should be, but I would guess that the claim is false in general. This is partly because, as you said, this is a particular case of the compact-open topology on function spaces, which is known to be well-behaved for locally compact spaces (and some more general spaces), but not for all spaces.

The answer is yes if $X$ is locally compact. Indeed, consider the topological space $X \times \mathcal{K}$ and let $V = \{(x,U) \in X \times \mathcal{K} | x \in U\}$. We claim that $V$ is open in $X \times \mathcal{K}$. To see this, take $(x,U) \in V$ and let $W$ be an open neighborhood of $x$ such that $\overline{W}$ is compact and contained in $U$. Then the set $W \times (\mathcal{K} \cap \mathcal{V}_{\overline{W}}) \subseteq X \times \mathcal{K}$ is an open neighborhood of $(x,U)$ which is contained in $V$, and so $V$ is open. Let us now show that $\cap_{U \in \mathcal{K}} U \subseteq X$ is open. First observe that $y \in \cap_{U \in \mathcal{K}} U$ if and only if $y \times \mathcal{K}$ is contained in $V$. Since $V$ is open and $\mathcal{K}$ is compact, if $y \times \mathcal{K} \subseteq V$ then there exists an open neighborhood $U_y$ of $y$ such that $U_y \times \mathcal{K} \subseteq V$ (this is a standard compactness argument). We may finally conclude that $U_y \subseteq \cap_{U \in \mathcal{K}} U$ and so $\cap_{U \in \mathcal{K}} U$ is open.

If $X$ is not locally compact then I'm not sure what the answer should be, but I would guess that the claim is false in general. This is partly because, as you said, this is a particular case of the compact-open topology on function spaces, which is known to be well-behaved for locally compact spaces (and some more general spaces), but not for all spaces.

Edit: Here is a template counter-example which shows that the claim is not true for general topological spaces. Let $X$ be a $T_1$-space and $A \subseteq X$ a subset satisfying the following two properties:

1. The intersection of $A$ with any compact subset of $X$ is finite.

2. $A$ is not closed.

Let us set $U_a = X \backslash \{a\}$. Then the collection $\mathcal{K} = \{U_a\}_{a \in A}$ is compact in the compact-open topology because by (1) every open set in $\mathcal{K}$ is cofinite. On the other hand, $\cap_{U \in \mathcal{K}}U = X \backslash A$ is not open by (2). To show that such spaces exist choose a non-principal ultrafilter $\mathcal{F}$ on the set $\mathbb{N}$ of natural numbers and set $X = \mathbb{N} \cup \{\infty\}$ where the open neighborhoods of $\infty$ are given by $\infty \cup U$ for $U \in \mathcal{F}$. Let $A = \mathbb{N} \subseteq X$. Then it is not hard to verify that $A$ satisfies properties (1) and (2) above.