Proof of Riesz's lemma

Proof

We take a point outside $Y$, move it by a nearly closest point of $Y$, and normalise.

1. Since $Y \neq X$, choose $x_0 \in X$ with $x_0 \notin Y$. Let $r = \inf_{y \in Y} \|x_0 - y\|$. This is a real number $r \ge 0$, since $Y$ contains $0$ and the norms are nonnegative.
2. $r > 0$. If $r = 0$, then for each $\varepsilon > 0$ there is $y \in Y$ with $\|x_0 - y\| < \varepsilon$. So $x_0$ is in the closure of $Y$, by closure in a metric space.
3. As $Y$ is closed, it equals its closure, by this result. So $x_0 \in Y$, a contradiction.
4. Since $0 < \alpha < 1$ and $r > 0$, we have $r / \alpha > r$. So $r / \alpha$ is not a lower bound of the distances, and there is $y_0 \in Y$ with $\|x_0 - y_0\| < r / \alpha$.
5. Let $d = \|x_0 - y_0\|$. Then $d \ge r > 0$. Let $x = (x_0 - y_0) / d$. Then $\|x\| = 1$.
6. Let $y \in Y$. Then $x - y = (x_0 - y_1) / d$, where $y_1 = y_0 + d \, y$.
7. As $Y$ is a linear subspace, $y_1 \in Y$. So $\|x_0 - y_1\| \ge r$, by the definition of $r$.
8. Hence $\|x - y\| = \|x_0 - y_1\| / d \ge r / d > r / (r / \alpha) = \alpha$.
9. So $\|x\| = 1$ and $\|x - y\| \ge \alpha$ for all $y \in Y$.