Let $X$ be a Banach space, and let $M,N$ be finite dimensional subspaces of $X$ such that $\dim M > \dim N$. Prove, that there exists $x\in M$ with $\Vert x\Vert = 1 = dist (x,N)$.
I am beginner in approximation theory so I need some advices. I found similar problem, more or less, and there was a hint: to take any $x_{0}\in X\setminus E$ and consider $x= d(x_{0},E)^{-1}(x_{0} - P_{E}(x_{0}))$, where $P_{E}(x)$ is the set of all $e\in E$ such that $d(x,E)=d(x,e)$. But there X was just infinite dimension space (not Banach) and $E$ just finite dimensional subspace, so maybe this is not useful in my case.