Let $W$ be a strict subspace of a Banach space $X$ and let the distance between a vector $v$ and the subspace $W$ be defined as $d(v,W)=\inf_{w\in W}|w-v|$.
If $X$ is a Hilbert space, then we can always find a unit vector $v$ st. $d(W,v)=1$. Does this also hold for Banach spaces? If not, what is a counter example?
This does not hold for Banach spaces in general. For example, take a look at this (notice that $X$ is not reflexive.)
However, it is true for reflexive Banach spaces:
Let $X$ be a reflexive Banach space (i.e. the closed unit ball of $X$ is weakly compact) and let $W$ be a closed subspace of $X$. Take any $x \in X \setminus W$. Pick a sequence $(w_n) \subset W$ such that $ ||x-w_n|| \downarrow d(x,W) \ $ (by the definition of the infimum). Since $||w_n|| \leq ||x|| + ||x-w_n|| $ we see that $(w_n)$ is bounded. By the EBerlein - Smulian theorem, we infer that $(w_n) \subset W $ has a weakly convergent subsequence. Hence, we may assume that $ w_n \xrightarrow{w} w $ for some $w \in W$ (notice that $W$ is weakly closed (by Mazur's theorem) and thus $w \in W$.) By the weak lower semi continuity of the norm $ || \cdot || \colon (X,w) \to \mathbb R$ we get that $$ d(x,W) \leq ||x-w|| \leq \liminf ||x_n -w|| = d(x,W) $$ and thus $d(x,W)=||x-w||$. We essentially showed that every closed subspace of $X$ is a proximinal set (the same proof applies for closed convex subsets). Now, since $d(x,W) = d(x-w,W) $ we conclude that $$d \left (\frac{x-w}{||x-w||} ,W \right)=1 $$
The replacement for the Hilbert space result is given by Riesz Lemma: https://en.wikipedia.org/wiki/Riesz%27s_lemma
