$\mathrm{dim}(\partial U_k)\leq n$ for all $k$, implies $\mathrm{dim}(\partial (\bigcap_k U_k))\leq n$? Let $X$ be a locally compact, Hausdorff, second countable space.
Let $(U_k)_{k\in\mathbb{N}}$ be a countable family of closed subsets in $X$ with $\mathrm{dim}(\partial_X U_k)\leq n$, for all $k$ (where $\mathrm{dim}$ stands for topological covering dimension). Does it follow that $\mathrm{dim}(\partial_X (\bigcap_{k\in\mathbb{N}} U_k))\leq n ?$
If the collection was finite, then the answer would be 'yes', as
$\partial_X (\bigcap\limits_{k=1}^{m} U_k)\subseteq \bigcup\limits_{k=1}^{m} \partial_X U_k$.
It is not clear that a similar inclusion holds for infinite collections, as here Boundary, unions and intersections I think that there is a mistake in the answer, the inclusion $\mathrm{int}( \bigcap \limits_{A \in \mathscr{A}} A)\subseteq \bigcap \limits_{A \in \mathscr{A}} \mathrm{int}(A)$ is reversed.
If the answer is 'no', I would be still happy to know about a class of topological spaces in which this holds (namely, having more restrictions on $X$).
 A: I assume that $\partial_X A$ is a boundary $\overline{A}\setminus\operatorname{int} A$ of a subset $A$ of the space $X$. 
Unfortunately, there is a counterexample even for $n=0$. For instance, let $X$ be a Cartesian product of a convergent sequence $\{0\}\cup\{1/m: m\in\Bbb N\}$ with a unit segment $[0,1]$ and $U_k=(\{0\}\cup\{1/m:  m\in\Bbb N\mbox{ and }m\ge k\})\times [0,1]\subset X$ for each $k$.  Then $\partial_X U_k=\emptyset$ for each $k$, but $\partial_X (\bigcap_{k\in\mathbb{N}} U_k)= \bigcap_{k\in\mathbb{N}} U_k=\{0\}\times [0,1]$ has dimension $1$.

If the answer is 'no', I would be still happy to know about a class of topological spaces in which this holds (namely, having more restrictions on X).

I expect that this holds for $X=\Bbb R^d$. Indeed, put $U=\bigcap_{k\in\mathbb{N}} U_k$. If some $U_k$ is nowhere dense, then $\partial_X U_k=U_k\supset U\supset \partial_X U$. 
If each $U_k$ equals $\Bbb R^d$ then the claim is trivially true. 
If for some $k$, $U_k\ne\Bbb R^d$ then since $\partial_X U_k\supset \partial_X\operatorname{int} U_k$, Exersise 7.4.18 from [Eng] implies that $$d-1\ge \dim (\partial_X U_k)\ge \dim (\partial_X\operatorname{int} U_k)=d-1,$$ so $n\ge d-1$. On the other hand, since $\partial_X U$ is nowhere dense, $\dim \partial_X U\le d-1\le n$. 

I guess the claim also holds when $X$ is a second countable d-manifold, because for it still holds 
$\dim A\le d-1$ for each a closed nowhere dense subset $A$ of $X$, see [Corollary 20, Sch].
References
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
[Sch] Reinhard Schultz, Notes on Topological Dimension Theory.
