Lemma: Let $T: \mathcal D(T) \to Y$ be a bounded linear operator with domain $\mathcal D(T) \subset X$, where $X$ and $Y$ are normed spaces. If $T$ is closed and $Y$ is complete, then $\mathcal D(T)$ is closed.
We are given the following proof:
Suppose that $T:\mathcal D(T) \to Y$ is a bounded, closed linear operator and that $Y$ is complete. We must show that $\mathcal D(T)$ is closed in $X$.
Let $(x_n)$ be a sequence in $\mathcal D(T)$ with $x_n \to x \in X$. We must show that $x \in \mathcal D(T)$.
Since $T$ is bounded, $$||Tx_n - Tx_m|| = ||T(x_n - x_m)||\le \|T\|\|x_n-x_m\|$$ for all $n, m \in \mathbb N$. Since $x_n \to x$, $(x_n)$ is Cauchy in $X$. Therefore $(Tx_n)$ is Cauchy in $Y$. Since $Y$ is complete, $Tx_n \to y \in Y$ . Since $T$ is closed, $x \in \mathcal D(T)$ and $y=Tx$. Therefore $\mathcal D(T)$ is closed in $X$.
My question, however, is: Do we really need the completeness of $Y$ to deduce that $\mathcal D(T)$ is closed? We did not use the completeness of $Y$ to show that $x \in \mathcal D(T)$. We only needed completeness to show that $y=Tx$, which, if I'm not mistaken, is not really necessary to show that $x \in \mathcal D(T)$.
I understand there is the following version of the Closed Graph Theorem:
Let $T: \mathcal D(T) \to Y$ be a linear operator, where $\mathcal D(T)\subset X$ and $X$ and $Y$ are normed spaces. Then $T$ is closed if and only if whenever $x_n \to x$ where $x_n \in \mathcal D(T)$ and $Tx_n \to y$, then $x \in \mathcal D(T)$ and $y = Tx$.
So, taking the abovementioned theorem into consideration, we know that since $T$ is closed, that $x_n \to x$ with $x_n \in \mathcal D(T)$, we must have that $x \in \mathcal D(T)$, which already proves the closedness of $\mathcal D(T)$. The additional information that $y=Tx$ does not really seem to be necessary to conclude the closedness of $\mathcal D(T)$ in $X$.
So, if this is the case, then why have the additional assumption in the original question that $Y$ is complete, if we don't really need it?
EDIT:
My trail of thinking of proving closedness of $\mathcal D(T)$ without needing the completeness assumption would be something as follows:
Suppose that $T: \mathcal D(T) \to Y$ is a closed, bounded linear operator. We must show that $\mathcal D(T)$ is closed in $X$.
Let $(x_n)$ be a sequence in $\mathcal D(T)$ with $x_n \to x$. We must show that $x \in \mathcal D(T)$. However, since $T$ is closed, we know that $x \in \mathcal D(T)$, i.e. $\mathcal D(T)$ is closed in $X$.
This uses the exact same argument as the first proof, however, it focuses only on the part necessary to show that $x \in D(T)$ and not that $y=Tx$ if $Tx_n \to y$, making it possible to omit the assumption that $Y$ is complete (that is, if I'm not incorrect, which is very possible).
Counterexample: Let $X$ be a Banach space and $Y$ a dense linear subspace that is not all of $X$. Take $T$ to be the identity map $\mathcal D(T) \to Y$ where $\mathcal D(T) = Y$.