Show $\pi$ is the orthogonal projection of $W$ iff $\|\pi(u)\| \leq \|u\|$ for all $u \in V$

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Let $V$ be an inner product space and $W \subseteq V$ a finite-dimensional subspace of $V$. Let $\pi \in \mathcal L(V,V)$ a projection with $W$ as image. Show that $\pi$ is the orthogonal projection $\operatorname{pr}_W$ of $W$ if and only if $\|\pi(u)\| \leq \|u\|$ for all $u \in V$.

I prove the first part $(\Rightarrow)$, but I can't figure out how to start the second part $(\Leftarrow)$.

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Suppose $w \in W^\perp$ such that $\pi(w) \neq 0$ and consider $u = tw + \pi(w)$ for real scalar $t$. Then $$\|u\|^2 = t^2\|w\|^2 + \|\pi(w)\|^2 \\ \|\pi(u)\|^2 = (t+1)^2\|\pi(w)\|^2$$ Then solve for $t$ such that $\|\pi(u)\| > \|u\|$