a one-to-one and onto function that cannot be continuous I'm reading Torchinsky's Real Analysis, and am having difficulty demonstrating   if
$$\varphi : [0,1]^2 \rightarrow [0,1]$$ is one-to-one and onto, then it cannot be continuous. (5.13)
Unable to demonstrate this myself, I did some research and found what seems to a proof: 
"Suppose such a function exists $$\varphi : [0,1]^2 \rightarrow [0,1]$$ 
Now, $[0,1]^2 \setminus\{\varphi^{-1}(1/2)\}$ is connected (since $\varphi$ is a bijection we are just removing one point from the square) and by properties of continuous functions we know that $\varphi$ maps connected sets to connected sets (cf. Rudin 4.22) . But $$\varphi\left([0,1]^2 \setminus\{\varphi^{-1}(1/2)\}\right) = [0, 1/2)\cup(1/2, 1]$$ which is not connected, hence no such map can exist."
What is unclear to me: removing point from the domain introduces a discontinuity in the codomain since the function is onto, but how does this prove the assertion for the domain in question: $$ [0,1]^2 \text{ ?}$$

We assumed the existence of such a function $\varphi$, did some work to arrive at a contradiction (in particular, the existence of a continuous map that maps a connected set to a set that is not connected). Therefore the assumption was not consistent.
The author might have meant $\varphi$ instead of $\varphi^{-1}$. That might have caused confusion.
