Prove that if $f:A\to B$ is uniformly continuous on $A$ and $g$ is uniformly continuous on $B$, then $g(f(x))$ is uniformly continuous on $A$ Suppose that $f\colon A \to B$ is uniformly continuous on $A$ and $g$ is uniformly continuous on $B$. Show that $g \circ f$ is uniformly continuous on $A$.
I tried to use the definition of uniformly continuous but it doesn't work.

So is this the usual "composition of uniformly continuous functions is uniformly continuous"? 
The uniform continuity of $f$ means that $\forall \epsilon_1 >0$ there exists $\delta_1 >0$ such that $|x-y| < \delta_1 $ implies $|f(x)-f(y)|< \epsilon_1$. Analogously, the uniform continuity of $g$ means that $\forall \epsilon_2 >0$ there exists $\delta_2 >0$ such that $|w-z| < \delta_2 $ implies $|g(w)-g(z)|< \epsilon_2$. So, once you have chosen $\epsilon_2$, you can find $\delta_2$ as prescribed, and then find a suitable $\delta_1$ accordingly, by imposing $\epsilon_1 \leq \delta_2$, thus obtaining $|g \circ f(x)-g \circ f(y)|< \epsilon_2$ whenever $|x-y| < \delta_1$. It's sort of like nesting two $\epsilon - \delta$ arguments. This is morally the thesis, just write it down a bit more formally if you need to. 
