Proof of The Distance Between the Values of Two Continuous Maps Is Continuous

A bound for the differences of $h$. Let $s,t\in A$. By Metric Space (v), applied twice,

$$h(s)=d_{Y}(x(s),y(s))\le d_{Y}(x(s),x(t))+d_{Y}(x(t),y(t))+d_{Y}(y(t),y(s)),$$

and, with $s$ and $t$ exchanged, $h(t)\le d_{Y}(x(t),x(s))+h(s)+d_{Y}(y(s),y(t))$. By Metric Space (iv), $d_{Y}(x(t),x(s))=d_{Y}(x(s),x(t))$ and $d_{Y}(y(t),y(s))=d_{Y}(y(s),y(t))$, so both $h(s)-h(t)$ and $h(t)-h(s)$ are at most $d_{Y}(x(s),x(t))+d_{Y}(y(s),y(t))$, that is (the difference and the absolute value being read in the type $\mathrm{Real}$ of The Real Numbers as a Type (i), by Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (i) and Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (iii)),

$$d_{\mathbb{R}}(h(s),h(t))=|h(s)-h(t)|\le d_{Y}(x(s),x(t))+d_{Y}(y(s),y(t)).\qquad(1)$$

Continuity ((i)). Let $t\in A$. The choices are made in this order. First a real $\varepsilon>0$ is given; its half $\varepsilon/2$ is a real number with $\varepsilon/2+\varepsilon/2=\varepsilon$ by Closure Facts in an Ordered Field: Negatives, Differences, Quotients, Absolute Values, Powers, Natural Numbers, Halves and Minima (vii), with the ordered field $\mathbb{R}$ for its $F$. Then, as $x$ is continuous on $A$, it is continuous at $t$ relative to $A$ by Continuous Map Between Metric Spaces (ii), and Continuous Map Between Metric Spaces (i), applied to $\varepsilon/2>0$, gives a real $\delta_{1}>0$ such that every $s\in A$ with $d_{X}(t,s)<\delta_{1}$ satisfies $d_{Y}(x(s),x(t))<\varepsilon/2$. In the same way, continuity of $y$ at $t$ relative to $A$, applied to $\varepsilon/2$, gives a real $\delta_{2}>0$ such that every $s\in A$ with $d_{X}(t,s)<\delta_{2}$ satisfies $d_{Y}(y(s),y(t))<\varepsilon/2$. Let $\delta>0$ be the smaller of $\delta_{1}$ and $\delta_{2}$; it depends on $t$ and $\varepsilon$ only. Now let $s\in A$ with $d_{X}(t,s)<\delta$. Then $d_{X}(t,s)<\delta_{1}$ and $d_{X}(t,s)<\delta_{2}$, and by (1)

$$d_{\mathbb{R}}(h(s),h(t))\le d_{Y}(x(s),x(t))+d_{Y}(y(s),y(t))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

So $h$ is continuous at $t$ relative to $A$, as a map from $(X,d_{X})$ to $(\mathbb{R},d_{\mathbb{R}})$. As $t\in A$ was arbitrary, $h$ is continuous on $A$ by Continuous Map Between Metric Spaces (ii).