A fixed point for every continuous self-map of [0,1]

Idea. A fixed point is a point where the graph of $f$ meets the diagonal $y=x$. The graph starts above (or on) the diagonal and ends below (or on) it, so it must cross.

Proof. Let $g(x)=f(x)-x$, which is continuous on $[0,1]$. Since $f$ takes values in $[0,1]$:
- $g(0)=f(0)\ge0$,
- $g(1)=f(1)-1\le0$.

If $g(0)=0$ or $g(1)=0$, we are done. Otherwise $g(0)>0>g(1)$ and the intermediate value theorem gives $c\in(0,1)$ with $g(c)=0$, that is $f(c)=c$.

Remark. Both hypotheses matter. Continuity: $f(x)=1$ on $[0,\frac12)$ and $f(x)=0$ on $[\frac12,1]$ has no fixed point. The interval being closed: $f(x)=\frac x2$ maps $(0,1]$ to itself with no fixed point. This is the one-dimensional case of Brouwer's fixed point theorem.