This is for the special case of functions from and into the unit square ($I^2$) and is not mine but a friend came up with it.
Considering the function $f(x,y)$ defined on the unit square, we define the function $g(x,y) = f(x,y)-(x,y)$. Now, if a point $P$ is on the line segment $(0,0)$ to $(0,1)$ then the $x$-coordinate of $g(P)$ must be greater than or equal to $0$, since the $x$-coordinate of $f(P)$ must be greater than or equal to $0$.
Similarly, if a point $P$ on the line segment $(1,0)$ to $(1,1)$ then $g(P)$ must have an $x$-coordinate less than or equal to $0$.
If we colour all points $P$ in the unit square such that the $x$-coordinate of $g(P)$ is positive red, all those where it is negative blue, and those where it is $0$ green, then the unit square can be divided into regions.
The leftmost edge of the unit square is red and/or green, and the rightmost is blue and/or green. Extend the unit square on the left and colour that area red. Then the red region must have a boundary within the square, and as $(0,0),(0,1)$ are either directly on that boundary or included within the red region, this boundary goes from top to bottom of the square.
This boundary must be green, as $f$ is continuous. Therefore, there is a path going the top to the bottom of the unit square whose image in $g$ has only points with $x$-coordinate $0$.
Similarly, we have a path from the left and right sides of $I^2$ whose image in $g$ has points of $y$-coordinate $0$. Thus, there is a point of intersection between these paths and at this point $P$, $g(P) = (0,0) = f(x,y)-(x,y)$ which is a fixed-point.
"QED" (maybe?)
Is this proof correct? I am aware it needs the introduction of some more rigour and further, if it is wrong, can this be fixed?
EDIT: Added colouring argument.