The thing is, homotopy theory is the two-dimensional version of the Intermediate Value Theorem. Here is the Intermediate Value Theorem:
Intermediate Value Theorem. Let $[a,b]$ be a closed interval. Let $f\colon [a,b] \to \mathbb{R}$ be a continuous function, and suppose that $f(a)$ and $f(b)$ lie on opposite sides of a point $q\in\mathbb{R}$. Then there exists a point $p\in [a,b]$ so that $f(p) = q$.
And here is the basic generalization of the Intermediate Value Theorem for functions in two dimensions:
Intermediate Value Theorem in Two Dimensions. Let $D$ be a closed disk, and let $\partial D$ denote its boundary circle. Let $f\colon D \to \mathbb{R}^2$ be a continuous function, and suppose that $f(\partial D)$ has nonzero winding number around a point $q\in\mathbb{R}^2$. Then there exists a point $p\in D\,$ so that $f(p) = q$.
A few notes:
In case you're not familiar with the concept, the winding number of a closed curve around a point $q$ in the plane is the number of times that the curve winds counterclockwise around the point. That is, the winding number is the element of $\pi_1(\mathbb{R}^2-\{q\})$ represented by the curve.
Here the requirement that $f(\partial D)$ winds around $q\,$ replaces the IVT hypothesis that the images of the endpoints of the interval lie on opposite sides of $q$. In both cases, the idea is that the image of the boundary somehow “surrounds” $q$.
There is even a generalization of the standard proof of the Intermediate Value Theorem to multiple dimensions.
Sketch of Proof of IVT: Suppose we bisect $[a,b]$ into two subintervals. Then one of these must also have the property that the images of its endpoints surround $q$. By repeatedly bisecting, we obtain a nested sequence of closed intervals whose intersection is a single point $p$. By continuity, $f$ must map $p$ to $q$.$\qquad\Box$
Sketch of Proof of IVT in Two Dimensions: Imagine the domain disk $D$ as a square, and imagine that we cut this square into four subsquares. Then the winding number of the image of the boundary of $D$ around $q\,$ is the sum of the winding numbers of the images of the boundaries of each of the four subsquares around $q\,$. In particular, if the winding number of $f(\partial D)$ around $q\,$ is nonzero, then the same must hold for at least one of the subsquares. Iterating this process, we obtain a nested sequence of closed squares, whose intersection must be a single point $p$, and it is easy to show that $f(p)$ must be $q$.$\qquad\Box$
Note that this proof contains an implicit algorithm for finding the root of the function $f\colon D \to \mathbb{R}^2$. This algorithm is a bit harder than the bisection algorithm in one dimension, since it involves computing some winding numbers.
Of course, none of this involves drawing the graph of $f\,$ in four dimensions, and for good reason: drawing the graph isn't even a particularly good way to prove the Intermediate Value Theorem in one dimension. Indeed, the only way to use graphs to prove the Intermediate Value Theorem is to invoke the much harder Jordan Curve Theorem. This method depends on the following result:
Path Intersection Theorem. Let $D$ be a closed disk, and let $\alpha\colon [0,1]\to D$ and $\beta\colon [0,1]\to D$ be continuous paths whose endpoints lie on $\partial D$. If the endpoints of $\beta\,$ lie on different components of $\partial D - \{\alpha(0),\alpha(1)\}$, then $\alpha$ and $\beta$ must intersect.
This theorem can be proven in a simple way using the Jordan Curve Theorem. It should be obvious how this can be used to prove the IVT.
Here is an analogous geometric theorem in four dimensions:
Disk Intersection Theorem. Let $D^4$ be a closed $4$-ball, and let $D^2$ be the closed unit disk. Let $\alpha\colon D^2 \to D^4$ and $\beta\colon D^2 \to D^4$ be continuous functions, and suppose that $\alpha(\partial D^2) \subset \partial D^4$ and $\beta(\partial D^2) \subset \partial D^4$. If the curves $\alpha(\partial D^2)$ and $\beta(\partial D^2)$ have nonzero linking number in the $3$-sphere $\partial D^4$, then the images of $\alpha$ and $\beta$ must intersect in $D^4$.
It seems to me that this theorem is harder than the Brouwer Fixed Point Theorem, but it does contain the essential geometry that must be used to prove the Brouwer Fixed Point Theorem (or the Two-Dimensional Intermediate Value Theorem) if you want to use graphs.
The IVT is not the only ingredient here. The way the theorem works is by setting up this square:
where the line in the middle is $y = x$. A function from $[0, 1]$ to $[0, 1]$ that intersects this line will have a fixed point at the point of intersection.
The IVT kicks in when we have a function whose graph enters the top triangle and the bottom triangle at various points, e.g.
The fact is, by the IVT, the function has to cut the line somewhere, i.e. it must have a fixed point.
But, this makes an assumption! The function may only exist in one triangle or the other, but not in both. That is, why can we not have $f(x) > x$ for all $x$ or $f(x) < x$ for all $x$?
The picture above illustrates it. Due to the function having a full domain of $[0, 1]$, there's a squeeze happening. The green function is about as close as we can have to a function satisfying $f(x) > x$. Similarly, the red function attempts to have $f(x) < x$. But, both are pinched towards the diagonal line. The green function must have $f(1) = 1$, and the red function must have $f(0) = 0$.
This illustrates the necessity of defining all the way to $0$ and $1$. Removing either of these points means that the functions are not squeezed to a fixed point (we'd only ensure that $f(x)$ and $x$ become arbitrarily close).
Best Answer
Tsemo's answer is correct, but intuitively, it follows because when your interval is missing an endpoint, you can 'hide' the fixed point in the 'hole' formed by the missing endpoint. For example, $f(x) = x^2$, viewed as a map from $(0,1)$ to $(0,1)$ is fixed point free, though of course, this is just because it's fixed points are 0 and 1, which are excluded from your domain.