Distance of element of normed space to finite dimensional subspace is realised by element of subspace

Question

Let $A$ be a normed space and $B$ a finite dimensional subspace of $A$. Let $a\in A$, then we define the distance of $a$ to $B$ as $d(a,B)=\inf_{b\in B}||a-b||$.

How do I prove that there exists a $b_a\in B$ such that $d(a,B)=||a-b_a||$?

I have no idea where to begin this proof, so I would greatly appreciate a hint to start me off.

Answer 1

If $d(a,B) = \eta$, there exists a sequence such $(b_n)_n \in B^{\mathbb{N}}$ such that $||b_n - a|| \rightarrow \eta$. Therefore the sequence is a bounded sequence of a finite dimensional normed space, hence Bolzano-Weirstrass yields that there exists a subsequence $(b_{n_k})_k$ that converges to $b_l \in B$, which finishes the proof.