Why can we find $\psi_k(x_k) > 3/4$?

Question

I have a question about a detail in the following proof (more detail is included than required, I think):

Prop:

Let $X$ be a normed linear space and $W$ a finite dimensional subspace of $X^*$. Then there is a finite subset $F$ of $X$ for which $$\|\psi\|/2 \leq \max_{x \in F} \psi(x) \quad \forall \psi \in W$$

Proof:

Since $W$ is finite dimensional, its closed unit sphere $S^* = \{ \psi \in W | \|\psi \| = 1 \}$ is compact and therefore is totally bounded. Choose a finite subset $\{ \psi_1, ... \psi_K\}$ of $S^*$ for which $S^{*} \subset \cup_{k=1}^nB(\psi_k,1/2)$. For $1 \leq k \leq n$, choose a unit vector $x_k \in X$ such that $\psi_k(x_k) > 3/4$...(proof goes on)

Why can we find $\psi_k(x_k) > 3/4$?

Answer 1

By the Hahn-Banach theorem consequences, it is $$\|\psi\|=\sup\{|\psi(x)|: \|x\|_X=1\}$$ Since $\|\psi_k\|=1$, we can find a unit vector $y_k\in X$ such that $|\psi_k(y_k)|>3/4$. Rotating properly (specifically counter-clockwise and as much as the argument of $\psi_k(y_k)$ is), we get a unit vector $x_k$ such that $\psi_k(x_k)>3/4$.