Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$. 

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

**Motivation.** Suppose you study a space that appears naturally, and you manage to bijectively parametrize with a continuous parameter that takes values is a nice parameter space (such as $l^2$). If you discover that the parametrization isn't a homeomorphism, you may wonder if you have proved anything at all.

 **Remarks**  (to the original question):

1. $Y$ is path-connected; in fact, no countable (or finite) subset can separate $Y$.
(Here is why: the preimage of a countable subset of $Y$ is a countable subset of $X$.
Then it is known that the complement in $l^2$ of a countable subset is homeomorphic to $l^2$, and hence the image of the complement in $Y$ is path-connected.)
2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism.
In fact, there is a [result](http://www.springerlink.com.www.library.gatech.edu:2048/content/n587u1h31v224642/)  of Savkin that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent  [paper](http://www.jstor.org/pss/10.4169/000298910X521698)  by Creswell in Monthly.
3. The same question is interesting when $X$ is $\mathbb R^n$. (I briefly thought that the answer in this case follows from the invariance of domain, and Bill kindly corrected me in comments).