Consider a sequence of continuous functions $f_n:[a,b] \to \mathbb{R}$. If their graphs $G_n$ converge to the graph $G$ of a continuous function $f$ (in the Hausdorff metric $d_H$), prove that $f_n$ uniformly converges to $f$.
For given $\epsilon \gt 0$, there exists an $N \in \mathbb{N}$, such that $n \ge N \implies d_H(G_n, G) \lt \epsilon$. But I don't know how to proceed. The difficulty is that for $x \in [a,b]$, I can't conclude from $d_H(G_n, G) \lt \epsilon$ that $|f_n(x)-f(x)| \lt \epsilon$. All I can say is, there is an $x' \in [a,b]$, such that $\sqrt{|x-x'|^2+|f_n(x)-f(x')|^2} \lt \epsilon$.
Any hints on this?
You aren't using the fact that $f$ is continuous, and in particular uniformly continuous on $[a,b]$. Let $\epsilon > 0$ be given. We must show that there exists $N$ so that $n \ge N$ implies $|f_n(x) - f(x)| < \epsilon$.
Apply uniform continuity: choose $\delta > 0$ so that $x,x' \in [a,b]$ implies $|f(x) - f(x')| < \dfrac \epsilon 2$. Define $\theta = \min \left\{\dfrac \epsilon 2, \delta \right\}$. Now choose $N$ so that $n \ge N$ implies $d_H(G_n,G) < \theta$.
Let $x \in [a,b]$. Then there exists a point (as you point out) $x' \in [a,b]$ with the property that $\sqrt{|x-x'|^2 + |f_n(x) - f(x')|^2} < \theta$. Thus $$|x - x'| < \theta \implies |x - x'| < \delta \implies |f(x) - f(x')| < \frac \epsilon 2$$ and $$|f_n(x) - f(x')| < \theta \implies |f_n(x) - f(x')| < \frac \epsilon 2.$$ This gives you $|f_n(x) - f(x)| < \epsilon$.