Let $(X,d)$ be nonempty compact metric space and $f : X\to X$ be a function satisfying $d(f (x), f (y)) < d(x, y)$ for all distinct pair of points $x, x \in X$. Show that $f$ have a fixed point and this fixed point is unique.
Hint: Define the function $g : X \to R$ by $g(x) = d(f (x), x)$. Assume that $f (x) \neq x$ for all $x \in X$. Use compactness to show the existence of a point $a \in X$ such that $g(a) \leq g (x)$ for all $x \in X$. Deduce a contradiction by considering $x=f (a)$.
You can use the result that tells you that a continuous function on a compact set achieves a minimum (notice that $f$ is continuous since it is a strict contraction). This gives you the existence of $a$. You get a contradiction from choosing $x=f(a)$ because then $$g(a) = d(f(a),a) \leq d(f(f(a)), f(a)) = g(x)$$ which contradicts the fact that $f$ is a contraction (since $a \neq f(a)$).
Uniqueness follows easily by contradiction as well since if you assume $\exists x, y \in X$ fixed points you get (using the fact that $f$ is a contraction) $$d(f(x),f(y)) = d(x,y) < d(x,y).$$
