Fixed point and a derivative bounded by one

Existence. $g(x)=f(x)-x$ is continuous with $g(0)=f(0)\ge0$ and $g(1)=f(1)-1\le0$. If one of them is $0$ we are done; otherwise $g(0)>0>g(1)$ and the intermediate value theorem gives $c\in(0,1)$ with $g(c)=0$, i.e. $f(c)=c$.

Uniqueness. Suppose $u<v$ are two fixed points. By the mean value theorem there exists $\xi\in(u,v)$ with $f(v)-f(u)=f'(\xi)(v-u)$. Since $f(u)=u$ and $f(v)=v$, this reads $v-u=f'(\xi)(v-u)$, so $f'(\xi)=1$, contradicting $|f'|<1$.

Remark. The hypothesis can be weakened to "$|f'|\le k<1$": the map is then a *contraction*, and for any starting point $x_0$ the sequence $x_{n+1}=f(x_n)$ converges to the fixed point (Banach's fixed point theorem).