Brouwer's Fixed Point Theorem/One-Dimensional Version/Proof Using Connectedness

Proof
By Subset of Real Numbers is Interval iff Connected, $[a,b]$ is connected.

Suppose there is no fixed point.

Then $f(a)>a$ and $f(b)>b$.

Let $U = \{x \in [a,b] : f(x)x \}$.

Then $U$ and $V$ are open in $[a,b]$.

By assumption, $U\cup V = [a,b]$.

Thus $[a,b]$ is not connected, which is a contradiction.