[proofplan] We prove [uniform continuity](/page/Uniform%20Continuity) of the composition directly from the $\varepsilon$-$\delta$ definition. Given a tolerance in $Z$, uniform continuity of $g$ gives a tolerance in $Y$. Uniform continuity of $f$ then gives a single tolerance in $X$ that forces the images under $f$ to be close enough for the estimate for $g$ to apply. [/proofplan]

[step:Choose the intermediate tolerance from uniform continuity of $g$] Define the composition map \begin{align*} h:X\to Z,\qquad h(x)=g(f(x)) \text{ for every } x\in X. \end{align*} Let $\varepsilon>0$. Since $g:Y\to Z$ is uniformly continuous, there exists $\eta>0$ such that for all $y_1,y_2\in Y$, \begin{align*} d_Y(y_1,y_2)<\eta \implies d_Z(g(y_1),g(y_2))<\varepsilon. \end{align*} [/step]

[step:Use uniform continuity of $f$ with the intermediate tolerance]Since $\eta>0$ and $f:X\to Y$ is uniformly continuous, there exists $\delta>0$ such that for all $x_1,x_2\in X$, \begin{align*} d_X(x_1,x_2)<\delta \implies d_Y(f(x_1),f(x_2))<\eta. \end{align*}[/step]

[guided]The number $\eta$ is the amount of closeness in $Y$ required in order for $g$ to produce $\varepsilon$-closeness in $Z$. Because uniform continuity of $f$ works for every positive tolerance in $Y$, and because the previous step gives $\eta>0$, we may apply uniform continuity of $f$ with this specific tolerance $\eta$. Thus there exists $\delta>0$ such that for every pair $x_1,x_2\in X$, \begin{align*} d_X(x_1,x_2)<\delta \implies d_Y(f(x_1),f(x_2))<\eta. \end{align*} The important point is that this $\delta$ depends only on $\eta$, and hence ultimately only on $\varepsilon$, not on the particular points $x_1$ and $x_2$. This is exactly the uniform part of uniform continuity.[/guided]

[step:Combine the two estimates to prove uniform continuity of the composition] Let $x_1,x_2\in X$ satisfy $d_X(x_1,x_2)<\delta$. By the choice of $\delta$, \begin{align*} d_Y(f(x_1),f(x_2))<\eta. \end{align*} Applying the choice of $\eta$ with $y_1=f(x_1)$ and $y_2=f(x_2)$ gives \begin{align*} d_Z(g(f(x_1)),g(f(x_2)))<\varepsilon. \end{align*} Since $h(x)=g(f(x))$ for every $x\in X$, this says \begin{align*} d_Z(h(x_1),h(x_2))<\varepsilon. \end{align*} Therefore, for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x_1,x_2\in X$, \begin{align*} d_X(x_1,x_2)<\delta \implies d_Z(h(x_1),h(x_2))<\varepsilon. \end{align*} Hence $h=g\circ f:X\to Z$ is uniformly continuous. [/step]