Somewhat shady way of solving a problem from Baby Rudin

Question

The problem statement is: If $\mathbb{R}^n$ is the (countably) infinite union of closed sets, show at least one of those closed sets has non empty interior.

My shady way of solving this is noting that a closed set without interior is a boundary (i.e. of some open set) and therefore has n-dimensional Lebesgue measure 0. A countable union of null sets has measure zero, so it can't be $\mathbb{R}^n$.

Is this method correct (but obviously violating the spirit of the problem as an exercise in metric topology), fundamentally wrong, or is it circular?

Answer 1

Lebesgue measure of a boundary (of an open set) need not be zero. Not even in $\mathbb R^1$.

Answer 2

Suppose that $\mathbb{R}^n$ is the countable union of closed sets each of which has empty interior. That is, $\mathbb{R}^n = \cup_{k=1}^\infty C_k$ where each closed set $C_k$ is nowhere dense by our assumption. But then since $\mathbb{R^n}$ is complete, this contradicts the Baire Category theorem.