Brouwer's Fixed Point Theorem/One-Dimensional Version/Proof by Intermediate Value Theorem

Proof
As the codomain of $f$ is $\closedint a b$, it follows that the image of $f$ is a subset of $\closedint a b$.

Thus $\map f a \ge a$ and $\map f b \le b$.

Let us define the real function $g: \closedint a b \to \R$ by $\map g x = \map f x - x$.

Then by the Combined Sum Rule for Continuous Functions, $\map g x$ is continuous on $\closedint a b$.

But $\map g a \ge 0$ and $\map g b \le 0$.

By the Intermediate Value Theorem, $\exists \xi: \map g \xi = 0$.

Thus $\map f \xi = \xi$.