I'm studying an introductory metric topology book and the following exercise came up.
Prove the Riesz Lemma: Let $F$ be the closed vector subspace of the normed space $E$, $F \neq E$ and $0 < r < 1$. Show there exists $e \in E$ such that $|e| = 1$ and $d(e, F) > r$. Hint: Choose $x \in E \setminus F$, denote $t = d(x,F)$ and choose $y \in F$ such that $|y-x| < t/r$. Set $e = (x-y)/|x-y|$. Exercise 2:17 may be helpful.
Result of exercise 2:17
If $F$ is the vector subspace of the normed space $E$, $x \in E, y \in F, a >0$ then $d(y+ax,F) = ad(x,F)$.
My work so far: $|e| = |x-y|/|x-y| = 1$. Now $$d(e,F) = \text{inf}\{|e-a|:a\in F\} = \text{inf}\{|\frac{x}{|x-y|}-\frac{y}{|x-y|} - a|:a \in F\}$$ By the exercise 2:17 result $\inf\{|\frac{x}{|x-y|} - \frac{y}{|x-y|} - a|\} = \frac{1}{|x-y|}\inf\{|x-a|\}$ and by the assumptions $\frac{1}{|x-y|}\text{inf}\{|x-a|\} > \frac{1}{t/r}\text{inf}\{|x-a|\} = r$. Is my approach correct?
EDIT1: I'm also wondering why do we assume $F$ is closed?