Existence of a closest matrix of rank at most r

Closedness of $\mathcal{R}_r$. $\operatorname{rank}B\le r$ iff all minors of size $r+1$ vanish. Minors are polynomials in the entries, hence continuous, and $\mathcal{R}_r$ is an intersection of zero sets of continuous functions, so it is closed. (For $r=n$, $\mathcal{R}_r=M_n(\mathbb{R})$.)

Reduction to a compact set. $0\in\mathcal{R}_r$, so the minimum, if it exists, is at most $N(A)$. Consider
$$K=\mathcal{R}_r\cap\{B:N(B)\le2N(A)\}.$$
This is closed (closed $\cap$ closed ball) and bounded, so it is compact since $M_n(\mathbb{R})\cong\mathbb{R}^{n^2}$. If $B\in\mathcal{R}_r\setminus K$, then $N(B)>2N(A)$ and
$$N(A-B)\ge N(B)-N(A)>N(A)=N(A-0),$$
so such a $B$ is worse than $B=0\in K$.

Existence. $B\mapsto N(A-B)$ is continuous, so it attains its minimum on the nonempty compact set $K$, at some $B_0$. By the previous paragraph $B_0$ is a minimiser on all of $\mathcal{R}_r$.

Remark. Existence holds for every norm, but uniqueness may fail (for the operator norm there are many best rank-$r$ approximations when $\sigma_{r+1}=\sigma_{r+2}$ or even in general). For the Frobenius and operator norms the minimum is explicit: see the Eckart-Young problem. The statement is false for the *set of rank exactly $r$* matrices, which is not closed.