Suppose $(X,\ d)$ is a compact metric space and $f:X\to X$ is a continuous function with the property that $$d(f(x),\ f(y))\le\frac{1}{2}d(x,\ y)$$ $\forall x,\ y\in X$.
Show that $f$ has a unique fixed point.
Hint: Define $f^1=f$ and $f^{n+1}=f\circ f^n$. Consider the intersection of the sets $A_n=f^n(X)$.
On
ignoring the hint, here is the easiest proof: define $$ g(x) = d(x,f(x)). $$ This is continuous, the space is compact, so $g$ assumes its minimum at some $x_0.$ Assume that $g(x_0) = \delta > 0.$ Then $g(f(x_0)) = d(f(x_0), f(f(x_0)) \leq \frac{1}{2} d(x_0, f(x_0)) = \frac{1}{2} \delta < \delta.$ This contradicts the assumption that $\delta \neq 0.$ That is, $g(x_0) = 0$ and $x_0$ is a fixpoint
The key point here is that a compact metric space has finite diameter (see, e.g., Compact metric space: proof $\text{diam}(K)$). $A_n$ is a nested sequence of sets (because the range of $f^n$ is contained in the range of $f^{n+1}$) with shrinking diameter approaching zero (because the diameter of $f(S)$ is $\leq \frac12$ the diameter of $S$ for any set $S \subset X$). So its intersection is a single point $p$; since $f(p)$ must also be in the intersection, $p$ is a fixed point.