Visually this is true for a finite dimensional inner product space. But when will this hold in an infinite dimensional one? I proved one side like so : Let $z\in M^\perp$. For $w\in M$ $$|\langle z,z-w\rangle|\leq\|z\|\|z-w\|$$ or $$\|z\|^2\leq\|z\|\|z-w\|$$ or $$\|z\|\leq\|z-w\|$$ and this holds true for all $w\in M$. Thus taking infimum we get $$\|z\|\leq d(z,M)$$. The equality follows since $0\in M$.
When does the converse hold?
Nevermind I found what I was looking for. Suppose $\|z\|=d(z,M)$. If possible let $m\in M$ be a unit vector such that $\langle z,m\rangle=\alpha\neq 0$. Then if we define $w=\alpha m$ we see that $\|z-w\|^2=\|z\|^2-|\alpha|^2$.