Why must a normed space homeomorphic to a complete metric space be complete?

I don't know if this works, but some thoughts:

It is well known that a metric space $(X,d_X)$ can be homeomorphic to a complete metric space $(Y,d_Y)$ and the metric $d_X$ on $X$ is not complete. E.g. $X = (0,1)$ in the standard metric, compared with $Y = \mathbb{R}$ in the standard (complete) metric. However, for such spaces $X$ it is the case that they are a $G_\delta$ in their completion (they are what is known as topologically complete).

Every normed space $(X,\|\cdot\|)$ has a normed completion $(\hat{X},\|\cdot\|)$ in which $X$ embeds isometrically. So we know from the above that $X$ is then a $G_\delta$ in $\hat{X}$. We want to show that these spaces are actually equal.

So maybe it is possible to show that a dense $G_\delta$ linear subspace of a Banach space is the whole space?

Added (after comments by Sean Eberhard) This is indeed the case: $X$ is a dense $G_\delta$ in $\hat{X}$, so if $y \in \hat{X}\setminus X$, $y + X$ and $X$ are both dense $G_\delta$'s and their intersection is empty (otherwise $y$ would a difference of members of $X$ hence in $X$). This contradicts Baire's theorem, so such a $y$ cannot exist and $X = \hat{X}$, and thus is complete already.