I recently asked this question, and this question is in some sense a generalization of it.
Can there be a path-connected, non-compact topological space $X$ and a continuous bijection $f:X\to [0,1]^n$?
This is a different question for every $n\in\mathbb{N}$: I would be happy to hear a result for any $n$ (while naturally I'd be most happy for a solution that will settle the matter for all of them). Clearly, if there is such a space for $n=1$, then there is for every $n$, but I wouldn't be surprised if this is impossible for $n=1$ while possible for larger $n$.
For the record: if we remove either condition, path-connectedness or non-compactness, this becomes trivial. In the first case, just take $[0,1]^n\setminus\{(0,...,0)\}\cup \{(3,0,...,0)\}$, while in the second case you just take $X=[0,1]^n$.
EDIT. The problem was solved by Paul Frost for $n>1$. So it only remains to see if this is possible or not for $n=1$.
Edit 2. The question has now been resolved completely.
Yes for $n > 1$.
It suffices to consider $n=2$, then $f \times id : X \times I^{n-2} \to I^2 \times I^{n-2} = I^n$ is an example for $n > 2$.
Let us identify $I^2$ with the unit disk $D^2 \subset \mathbb C$ and let $X' = (0,1] \times I$. Define $$f' : X' \to D^2, f'(s,t) = e^{2\pi i s} t .$$ This is a continuous surjection. Its restriction to $$ X = (0,1] \times(0,1] \cup \{(1,0) \}$$ is a continuous bijection .