Proving that $\min \{ \|x-x_0\|:x\in F\} = \max \{ | \langle x_0,y \rangle| : y \in F^{\perp} , \|y\|=1 \}$, if $H$ is Hilbert, $F$ a closed subspace

Question

So, I need to prove that $\min \{ \|x-x_0\|:x\in F\} = \max \{ | \langle x_0,y \rangle| : y \in F^{\perp} , \|y\|=1 \}$, if $H$ is a Hilbert space, $F$ is a proper linear subspace of $H$ and $x_0 \in H \setminus F$.
First of all, I know that $x_0$ can be written as $x_0 = x_1 + x_2$, $x_1 \in F$ and $x_2 \in F^{\perp} $, in a unique way, where $\|x_1-x_0\| = \min \{ \|x-x_0\|:x\in F\}$. I have proven that $|\langle x_0,y \rangle| \leq \|x_1-x_0\|$, so $\max \{ | \langle x_0,y \rangle| : y \in F^{\perp} , \|y\|=1 \} \leq \min \{ \|x-x_0\|:x\in F\}$.
The problem is I can't prove that $\min \{ \|x-x_0\|:x\in F\} \leq \max \{ | \langle x_0,y \rangle| : y \in F^{\perp} , \|y\|=1 \} $.
Any help would be appreciated.