Why must a normed space homeomorphic to a complete metric space be complete? Why must a normed space $X$ homeomorphic to a complete metric space $Y$ be complete?
I've solved this in the case where Y is a normed space (considering open balls gives Lipschitz equivalence), but am not sure how to proceed in general. Can we use an adapted open-balls argument even when Y isn't normed?
Maybe there's an equivalent condition for completeness that will come in useful? In a normed space we can use (absolutely convergent)$\implies$(convergent) for series, but this won't work in a metric space.
I've tried using an explicit homeomorphism $f:X\rightarrow Y$ and taking pullbacks and pushforwards along $f$ of a Cauchy sequence in $X$, but without success.
Many thanks for any help with this!
 A: 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.  
A: This question is not true:
$\mathbb{R}$ is homeomorphic to $(0,1)$. $\mathbb{R}$ is a complete metric space whit Euclidean norm and $(0,1)$ is a normed vector space which isn't complete!
