Answering this, as a new duplicate was recently closed.
I was trying to construct an as elementary proof as possible but...
I think that in order to use the fact that the inverse would also have to be continuous, we will have to use the fact that both domain and range are compact and Hausdorff, together with the respective result on continuity of the inverse of a bijection (see robjohn's answer).
Maybe the above can still be obtained in a more elementary way, so let us proceed:
Assume that we know the inverse $f^{-1}$ of such a function would be continuos, thus open subsets of $\left[0,1\right]$ would have an open preimage with respect to $f^{-1}$:
Specifically the image $C\;:=\;f\left(\left(0,1\right)\right)$ would be open in $B\;:=\;\left[0,1\right]\times\left[0,1\right]$.
The complement of $\left(0,1\right)$ in $\left[0,1\right]$ only contains the two points $0$ and $1$. These would have to be bijectively mapped to the complement of $C$ in $B$ which however contains more than $2$ points. Contradiction.
EDIT: I think I came about what could qualify as an "elementary" proof:
Consider the preimage of $J\,:=\,\left(0,1\right)\times\left(0,1\right)$. As $f$ is continuous, this preimage is open: $f^{-1}\left(J\right)=\left(a,b\right)\,=:\, I$ with $0<a<b<1$. The set $\left[0,1\right]\backslash I$ thus consists of two disjoint compact intervals, namely $A\,:=\,\left[0,a\right]$ and $B\,:=\,\left[b,1\right]$. As $f$ is supposed to be a continuous bijection, the set $K\,:=\,\left(\left[0,1\right]\times\left[0,1\right]\right)\backslash J$
is the disjoint union of the two images of $A$
and $B$
under the mapping $f$: $K=f\left(A\right)\,\biguplus\, f\left(B\right)$. However as $f$
is continuous, $f\left(A\right)$
and $f\left(B\right)$
are both connected and compact, thus such a partition of $K$
cannot exist. Contradiction.