Find a continuous bijection $f : X \to \mathbb{R}^2$ that isn't a homeomorphism, where $X$ is a specific subset of $\mathbb{R}^2$ 
Let $X$ be the subset $A \cup B$ of $\mathbb{R}^2$, where
  $$A = \{(x,y) \in \mathbb{R}^2 \mid x \le 0\text{ or }y \le 0\}$$
  and
  $$B = \{(x,y) \in \mathbb{R}^2 \mid x \lt 1\text{ and }y \lt 1\}.$$
Equip $\mathbb{R}^2$ with the standard product topology and equip $X$ with the induced subspace topology.
  Find a continuous bijection $f: X \to \mathbb{R}^2$ that is not a homeomorphism.

I don't really know what to do for this question, other than I think it needs to be a piecewise function. I know it can't simply be a translation of $B$ along the $x$ and $y$ axes, since that would have a continuous inverse function and hence be a homeomorphism. I think it somehow has to do with a reciprocal function and the fact that the $x = 0$ and $y = 0$ are a part of $A$ but are not contained in open intervals. 

Hint: Prove the quotient of $A\cup\overline B\setminus\{(1,1)\}$ given by $(t,1)\sim(0,\frac1{1-t})$ and $(1,t)\sim(\frac1{1-t},0)$, where $t\in(0,1)$ is homeomorphic to $\mathbb R^2$.
