If $H$ is a Hilbert space, does the coordinate projection $\pi :H\oplus H\rightarrow H$ take closed subspaces to closed subspaces? Here $H\oplus H$ has the product topology, which is induced from the "$\ell^2$-norm" $\|(x,y)\|:=\sqrt{\|x\|^2+\|y\|^2}$. This is indeed a Hilbert space via the inner product $\langle (x,y), (x',y')\rangle := \langle x,x'\rangle + \langle y,y'\rangle$.
Now we have a projection $\pi : H\oplus H \rightarrow H$, given by $(x,y)\mapsto x$. My question is: if $M$ is a closed subspace of $H\oplus H$, then is $\pi(M)$ a closed subspace of $H$?
 A: If this were the case, then any closed, densely defined operator on $H$ would automatically be bounded. Indeed, let $A$ be a closed, densely defined operator on $H$ with domain $D(A)$. By definition of a closed operator, the graph
$$\Gamma(A):=\{(x,Ax)\in H\oplus H\,:\,x\in D(A)\}$$
is a closed subspace of $H\oplus H$. If your question had a positive answer, this would imply $\pi(\Gamma(A))=D(A)$ is a closed subspace, so since $A$ is densely defined this means $D(A)=H$ and hence by the closed graph theorem $A$ is bounded.
In particular, we can find explicit counterexamples to your question by taking the graph of any closed, densely defined operator on a Hilbert space. For example,
$$H=L^2(0,1),\quad M=\{(u,v)\in H\oplus H\,:\,u\text{ is weakly differentiable and }u'=v\}.$$
A: No, this need not be the case. If you take $H := {\mathscr l}^2({\mathbb N})$, you can identify the projection $H\oplus H\to H$ with the map $\text{even}: {\mathscr l}^2({\mathbb N})\to {\mathscr l}^2({\mathbb N})$ mapping a sequence $(x_n)_n$ to the sequence $(x_{2n})_n$ of its even parts. Now, consider the subspace $U  := \{(x_n)\in {\mathscr l}^2({\mathbb N})\ |\ x_{2n+1} = n x_{2n}\text{ for all }n\in{\mathbb N}\}$. Then any eventually vanishing sequence belongs to $\text{even}(U)$ but $(1/n)_n\notin \text{even}(U)$. 
