Let $H$ be a Hilbert space and let $\pi \colon H \to H$ be an orthogonal projection. Let $E \subset H$ be a closed subspace of $H$.
My question: Is there any hope that one can conclude that $\pi(E)$ is closed?
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This is not true. Let's take $H=L^2(\mathbb R)$ and $E=PW_1$ as the subspace of all $f$ whose Fourier transform is supported by $[-1,1]$ (the Paley-Wiener space). Let $\pi$ be the projection onto $L^2(-1,1)$.
Then $f=\chi_{(-1,1)}(x)$ is in $\overline{\pi(E)}$: this follows because $$ \frac{\sin ax}{ax} = \frac{1}{2a}\int_{-a}^a e^{itx}\, dt \in E $$ for all small $a>0$, and clearly these functions converge to $f$ uniformly on $(-1,1)$ as $a\to 0$.
However, $f\notin\pi(E)$ because the functions from $E$ are entire, but the holomorphic continuation of $f$ from $(-1,1)$ to $\mathbb R$ is identically equal to $1$ and thus not in $L^2(\mathbb R)$.