I am trying to show any bijective mapping $f:I \to \mathcal{Q}$, where I is the unit interval in $\mathbb{R}$ and $\mathcal{Q}$ is the unit square, is necessarily discontinuous.
How do I go about proving this?
The context of this theorem is in the theory of space filling curves.
If $f$ continuous then $f^{-1}$ is also continuous: for any closed set $C$ in $\mathcal{Q}$ is compact and hence $f(C) = {({f}^{-1})}^{-1}(C)$ is compact and hence closed which is sufficient to prove the continuity of $f^{-1}$..
There is a $p \in \mathcal{Q}$ such that $f^{-1}(p) = 0.5$, the image of the connected set $\mathcal{Q} - \{p\}$ under $f^{-1}$ must be be connected, but it is $I - \{0.5\}$ which is not connected.
To see $\mathcal{Q} - \{p\}$ is path-connected, choose any two distinct points $p_1$ and $p_2$ in $\mathcal{Q}$, choose a third point $p_3 \in \mathcal{Q} - -\{p\}$ such that $p_1,p_2$ and $p_3$ form the vertices of a non degenerate triangle. $p$ can only lie on one of the sides of the triangle formed by these points so we can always move from $p_1$ to $p_2$ either along the edge joining them or by moving from $p_1$ to $p_3$ and then from $p_3$ to $p_2$ without leaving $\mathcal{Q} - \{p\}.$