Let $M_{1}$, $M_{2}$, and $M_{3}$ be metric spaces. Let $g$ be a uniformly continuous function from $M_{1}$ into $M_{2}$, and let $f$ be a uniformly continuous function from $M_{2}$ into $M_{3}$. Prove that $f(g(x))$ is uniformly continuous on $M_{1}$.
Solution
Let ε > 0. Since f is uniformly continuous, there exists some $δ > 0 $ such that $d2(x, y) < δ$ implies $d3(f(x),f(y)) < ε$ for all x,y ∈ M2. Since g is uniformly continuous, there exists some $γ > 0 $ such that d1(z,w) < γ implies $d2(g(z),g(w)) < δ $ for all z,w ∈ M1. Putting these together, we get that d1(z, w) < γ implies $d3(f(g(z))$, $f(g(w)) < ε $ for all z, w ∈ M1. Hence $f ( g (x))$is uniformly continuous on M1.
There is not enough explanation I think. I am waiting for your comments and extra ideas.