Riesz Lemma for reflexive spaces

Question

I know the proof of Riesz Lemma:

Let $Y$ be a closed (proper) subspace of a normed space $X$. Let $\varepsilon >0$. Then it exists an element $x \in X$ such that $||x||=1$ and $d(x, Y) \geq 1-\varepsilon$.

Now I want to face the case in which $X$ is reflexive. I read in this case $\varepsilon=0$.

To prove that, I think I have to use the following (James theorem?) but I don't manage to connect the dots...

$X$ is reflexive if and only if given $\Lambda \in X^*$ it exists $x \in X$ such that $||x||=1$ and $||\Lambda||=\Lambda x$.

I was thinking that $Y$ itself is also reflexive and applying the theorem to $Y$...

Answer 1

By Hahn-Banach, there is an $f\in X^*$ of norm 1 whose kernel contains $Y$.

In reflexive spaces, bounded linear functionals attain their norm of the closed unit sphere (this is the easy implication of James' theorem), so there is an $x\in X$ of norm 1 with $f(x)=1$.

So $1=f(x)=f(x-y)\le \Vert x-y\Vert$ for all $y\in Y$.