Assume not, then there exist $a<b$ with $f(a) - f(b) = \delta > 0$. By assumption, there are $c_1,c_2 \in [a,b],$ $c_1 \le c_2$ such that $$f(c_1)= \min_{x \in [a,b]} f(x) \le f(b) \qquad \text{and} \qquad f(c_2)= \max_{x \in [a,b]} f(x) \ge f(a).$$Now we must have $c_1 \le \frac{a+b}{2}$ and $f(a)-f(c_1) \ge \delta$, or $c_2 \ge \frac{a+b}{2}$ and $f(c_2)-f(b) \ge \delta$. In either case, there is an interval $[a_1, b_1] \subseteq [a,b]$ with $b_1 - a_1 \le \frac{b-a}{2}$ and $f(a_1) - f(b_1) \ge \delta$. Continuing in the same way, we find for every $n \in \mathbb{N}$ an interval $[a_n, b_n] \subseteq [a,b]$ with $b_n - a_n \le \frac{b-a}{2^n}$ and $f(a_n) - f(b_n) \ge \delta$. But this contradicts the fact that $f$ is uniformly continuous on $[a,b]$.