Lax Contains a proof of the following claim:
Claim: Let $Y$ be a closed, proper subspace of the normed linear space $X$. Then there is a vector $z$ in $X$ of length 1, $$\|z\| = 1$$ satisfying $$\|z-y\| >1/2$$ for all $y$ in $Y$.
Proof: Since $Y$ is a proper subspace of $X$, some point $x$ of $X$ does not belong to $Y$. Since $Y$ is closed, $x$ has a positive distance to $Y$: $$\inf_{y \in Y} \|x-y\| = d > 0$$ There is then a $y_0 \in Y$ such that $$\|x - y_0\| < 2d \quad (*)$$ Denote $z' = x - y_0$; we can then write $(*)$ as $\|z'\| < 2d$. It follows from definition of $d$ above that $$\|z' - y \| \geq d \quad (**)$$ for all $y$ in $Y$. Setting $z = \frac{z'}{\|z'\|}$ gives the vector and combining $(*)$ and $(**)$ gives the condition.
I'm failing to see how to put together $(*)$ and $(**)$ to finish the proof. Thoughts?
$\|z'-\|z'\|y\|\geq d$ since $\|z'\|y\in Y$, this implies that $\|{{z'-\|z'\|y}\over {\|z'\|}}\|\|\geq {d\over{\|z'\|}}$ Which implies that $\|{z'\over{\|z'\|}}-y\|\geq 1/2$