Proper subspaces are closed with empty interior

Idea. A subspace that contains even a tiny ball contains every direction, hence is everything. And finite-dimensional subspaces cannot have "holes" in their closure.

Step 1: $V$ is closed. $V$ is a finite-dimensional normed space, hence complete (it is isometric to some $\mathbb{R}^k$ after choosing an orthonormal basis for the Frobenius inner product). A complete subset of a metric space is closed: if $A_k\in V$ and $A_k\to A$, the sequence is Cauchy in $V$, so it converges to some $A'\in V$, and by uniqueness of limits $A=A'\in V$.

(Another argument: $V$ is the common kernel of finitely many linear functionals $\varphi_1,\dots,\varphi_m$, which are continuous, so $V=\bigcap\varphi_i^{-1}(\{0\})$ is closed.)

Step 2: empty interior. Suppose $V$ contained an open ball $B(A,r)$ with $A\in V$ (a point of the interior may be taken to be the centre of a ball inside $V$; since the ball is inside $V$, its centre is in $V$). Take any $X\neq0$ in $M_n(\mathbb{R})$ and put
$$t=\frac r{2\|X\|}.$$
Then $\|(A\pm tX)-A\|=r/2<r$, so $A+tX$ and $A-tX$ both lie in $B(A,r)\subset V$. Since $V$ is a subspace,
$$tX=\tfrac12\big((A+tX)-(A-tX)\big)\in V,\qquad\text{so}\qquad X=\tfrac1t(tX)\in V.$$
As $X$ was arbitrary, $V=M_n(\mathbb{R})$, contradicting properness.

Remark.
- The same proof works in any normed space: a proper subspace has empty interior. Closedness, however, genuinely needs finite dimension (the polynomials form a dense proper subspace of $C[0,1]$).
- This statement is used constantly: for instance the symmetric matrices, the trace-zero matrices and the commutant of a non-scalar matrix are all closed nowhere dense sets.