Let $K,G,H$ be Hilbert spaces with $D_A \subseteq K$, $D_B \subseteq G$ (possibly not dense) subspaces and let $A: D_A \rightarrow K$ and $B:D_B \rightarrow H$ be closed linear operators.
Then is the linear operator $B \circ A: A^{-1}(D_B) \rightarrow H$ a closed linear operator? where $A^{-1}(D_B)$ denotes the preimage of the domain of $B$.
Claim: This is the case when $D_A$ is complete and $A$ is bounded (see e.g. here). More generally, this is true when for any convergent sequence $(x_n)_{n \in \mathbb{N}}$ with $x_n \rightarrow x \in K$ we have $Ax_n \rightarrow y$ for some $y \in G$.
Proof of claim: Let $x_n \rightarrow x$ and $BAx_n \rightarrow y'$ for some $y' \in K$. Then by assumption $Ax_n \rightarrow y$. So since $A$ is closed, $x \in D_A$ and $Ax = y$. But now since $B$ is closed and $Ax_n \rightarrow y$ and $BAx_n \rightarrow y'$ we have $y \in D_B$ and $y' = By = B A x$, which proves the claim.
So I suspect that the answer to the original question is negative, but I have not been able to construct a counter example.
Partial answers and pointers to literature are very welcome. Thank you very much in advance.
Let $B:\ell^2(\Bbb N)\to\ell^2(\Bbb N)$ be given by $$B(\sum_{n} x_n \ e_n)=\sum_n 2^{-n}x_n \ e_n$$ this is a bounded operator hence closed. Further it is injective. Let $D=B(\ell^2(\Bbb N))$ be its image and $A:D\to \ell^2(\Bbb N)$ the left-inverse of $B$. $A$ is self-adjoint, and as such closed. The composition $B\circ A$ is just the inclusion $D\to \ell^2(\Bbb N)$, which is not closed in the sense of a densely defined operator on $\ell^2$.