This question is about the statement:
Let $X$, $Y$ be normed linear spaces and $D$ a linear subspace of $X$ and suppose that $A\colon D \to Y$ is a linear operator. If $A$ is continuous and closed and $Y$ complete, then $D$ is closed.
I was playing around a bit and tried to employ some techniques I had recently picked up in topology class in order to avoid tedious arguments with sequences (attempt at proof below):
Define $T: D\times D \to D \times Y$ by $T(x,x)=(x,Ax)$. Then $G_A=T(D\times D)$, where $G_A$ is the graph of $A$. Also, since $A, I$ are bounded, so is $T$ in the product norm and hence it is continuous. Thus $T^{-1}(G_A)=T^{-1}(T(D\times D))=D\times D$ is the inverse image of a closed under a continuous map and so is closed. Now $D=\Lambda^{-1}(D\times D)$ under the diagonal map $\Lambda$ which is continuous and hence $D$ is closed.
Now, I suspect that I messed up since I didn't use the completeness of $Y$ (and also I saw it proved in class in the standard way using completeness of $Y$). Could someone put me on the right track?
After some discussion it appears I was sloppy with defining the operator $T$ above. My counter-argument would then be to adjust as following (2nd attempt):
First define $U=\{(x,y) \in D\times D: x=y\}=G_I$. Since $U$is the graph of the identity it is a normed linear space as well. Now define $T:U \to D \times Y$ by $T(x,x)=(x,Ax)$. Then $G_A=T(U)$, where $G_A$ is the graph of $A$. Also, since $A, I$ are bounded, so is $T$ in the product norm and hence it is continuous. Thus $T^{-1}(G_A)=T^{-1}(T(U))=U$ is the inverse image of a closed under a continuous map and so is closed. Now $D=\Lambda^{-1}(U)$ under the diagonal map $\Lambda$ which is continuous and hence $D$ is closed.
Third attempt. I think I have found a construction which eliminates the issues highlighted by commenters and also uses the fact that $Y$ is a Banach space (thank you very much):
We begin with a lemma:
Let $X$ be a normed linear space and $Y$ a Banach space. If $A: D \subset X \to Y$ is a linear operator there exists a continuous extension of $A$ to all of $X$ such that $\hat A x= Ax$ for $x\in D$.
Define $S(x)=(x,Ax)$ and $\hat S(x)=(x,\hat Ax)$, where $\hat A$ is a continuous extension of $A$. Then $\hat S$ is a continuous extension of $S$. Note that $\hat Sx \in G_A \Rightarrow x\in D$. Now take $\hat S^{-1}(G_A)=D$. Hence $D$ is the inverse image of a closed set under a continuous function from $X$, so it is closed.