That was easy, but what if f is a continuous function from the unit disk into itself? It still has a fixed point, as proved by Brouwer. (<biography>)
Suppose f has no fixed point, and define g(x) as follows. For any x, draw a ray starting at f(x), through x, and exiting the unit circle at g(x).
One can write g(x) as an algebraic expression in terms of x and f(x), taking advantage of the quadratic formula as necessary. The formula for g(x), which I will not derive here, entails multiplication, division, addition, square root, and so on, and is continuous in x and f(x). Since f(x) is also continuous, g(x) is a continuous map from the disk onto the unit circle.
Note that g, restricted to S1, is the identity map with degree 1. Yet g is continuous throughout the unit disk, which implies a degree of 0 at the border. This is a contradiction, hence f admits a fixed point x, where f(x) = x.
If f maps the unit square into itself, deform the square into a disk and apply the above theorem. A continuous map from the unit square into itself has a fixed point. This holds for any closed region in the plane that is homeomorphic to the unit disk.
Of course a circle (without its interior) can rotate, so that there are no fixed points.
What about the open interval or the open disk? Consider the open interval (-1,1). For x ≥ 0, let x map to 1-½(1-x). For x ≤ 0, let x map to x + ½(x+1). The origin moves to the right, and squeezes the points ahead of it, and stretches the points behind it. this is a homeomorphism with no fixed points. Apply this (proportionally) to every horizontal chord in the open disk and find a continuous map with no fixed points.