Intersection of linear subspaces is $\{ 0 \}$ iff distance between respective unit vectors is positive in finite dimensions

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The question is from Linear Operator Theory in Engineering and Science by Arch W. Naylor and George R. Sell, Springer

Problem 5 from Section 5.10. Let $M$ and $N$ be linear subspaces of a normed linear space $X$ and $\delta(M,N)=\inf\{ \|x-y \|: x \in M, y \in N, \|x\|=\|y\|=1\}.$ Prove that if $X$ is finite dimensional, then $\delta(M,N)>0$ if and only if $M\bigcap N=\{0\}.$

I know to use that fact that in a finite dim. space, $B=\{x\in X:\|x\|\leq 1\}$ is compact. But I don't see where I can use it.

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Let $S=\{x\in X\mid\|x\|=1\}$. If $M\cap N\neq\{0\}$, take $v\in(M\cap N)\setminus\{0\}$. Then $\frac v{\lVert v\rVert}\in M\cap N$ and therefore $\delta(M,N)=0$.

If $M\cap N=\{0\}$, then $M\cap S$ and $N\cap S$ and disjoint closed subsets of $S$. Since $S$ is compact, $M\cap S$ and $N\cap S$ are compact too. And therefore $\delta(M,N)>0$, since $\delta(M,N)$ is the distance from $M\cap S$ to $N\cap S$, which is positive (since the sets are compact and disjoint).