Solution 1:

Let $g(x) = f(x) - x$. Then $g$ is continuous and $g(a) \geq 0$ while $g(b) \leq 0$. By the Intermediate Value Theorem, $g$ has at least one zero on $[a, b]$.

Solution 2:

For a different approach then the ones above, let us take $a = 0$ and $b = 1$. So assume $f:[0,1] \to [0,1]$ has no fixed point. Then $[0,1] = \{x \in [0,1] : f(x) < x\} \cup \{x \in [0,1] : f(x) > x \}$. Now argue that this is not possible.