TITLE: Riesz's Lemma for finite-dimensional spaces
QUESTION [6 upvotes]: Could someone give me an idea of ​​what happens in Riesz's lemma if the dimension of the space is finite?, i.e 
If $\dim Y<\infty$, then that one can even choose $\theta=1$. Why?

REPLY [5 votes]: This is a comment to David Mitra's answer. There is a beautiful result by Krein, Milman and Krasnoselski based on the Borsuk–Ulam theorem which strengthens the Riesz lemma in finite dimensions.

Theorem. Let $M$ and $N$ be finite-dimensional subspaces of a normed space. If $\dim M > \dim N$, then there is $x_0\in M$ such that ${\rm dist}(x_0, N)=\|x_0\|$.

I am happy to provide a proof on request. I am very curious whether there exists a proof of this statement which would avoid using the Borsuk–Ulam theorem.
Reference:

M. Kreĭn, M. Krasnoselski, D. Milman, On defect numbers of operators on Banach spaces and related geometric problems, Trudy Inst. Mat. Akad. Nauk Ukrain. SSR, 11 (1948), 97–112.