[proofplan] We prove [uniform continuity](/page/Uniform%20Continuity) directly from the $\varepsilon$-$\delta$ definition. The [discrete metric](/page/Discrete%20Metric) separates distinct points by distance $1$, so choosing a radius smaller than $1$ forces any two sufficiently close domain points to be equal. Once the two domain points are equal, their images under $f$ are equal, and the metric identity axiom in $Y$ gives distance $0$. [/proofplan] [step:Choose a uniform radius that forces equality in the discrete metric] Let $\varepsilon>0$ be arbitrary. Define \begin{align*} \delta &= \frac{1}{2} \end{align*} Then $\delta>0$. Let $x,x'\in X$ satisfy $d_X(x,x')<\delta$. Since $d_X$ is the discrete metric, its only possible values are $0$ and $1$, with $d_X(x,x')=1$ exactly when $x\neq x'$. The inequality $d_X(x,x')<\delta$ therefore excludes the case $x\neq x'$, so $x=x'$. [guided] We need a single choice of $\delta$ that works for every pair of points in $X$ and depends only on $\varepsilon$, not on the points themselves. In the discrete metric, distinct points are separated by the fixed distance $1$. Therefore any radius strictly smaller than $1$ has the useful property that two points within that radius cannot be distinct. Let $\varepsilon>0$ be arbitrary, and define $\delta:=\frac{1}{2}$. This is positive, so it is an admissible candidate in the definition of uniform continuity. Now let $x,x'\in X$ be any two points satisfying $d_X(x,x')<\delta$. Since $d_X$ is the discrete metric, we have $d_X(x,x')=0$ if $x=x'$ and $d_X(x,x')=1$ if $x\neq x'$. The inequality \begin{align*} d_X(x,x')<\frac{1}{2} \end{align*} is incompatible with $d_X(x,x')=1$. Hence the alternative $x\neq x'$ is impossible, and therefore $x=x'$. [/guided] [/step] [step:Use equality of nearby domain points to prove the uniform continuity inequality] From the previous step, every $x,x'\in X$ with $d_X(x,x')<\delta$ satisfies $x=x'$. Since $f:X\to Y$ is a map, this gives $f(x)=f(x')$. Since $d_Y$ is a metric on $Y$, the identity axiom gives \begin{align*} d_Y(f(x),f(x'))=d_Y(f(x),f(x))=0. \end{align*} Because $\varepsilon>0$, we have $0<\varepsilon$, and hence \begin{align*} d_Y(f(x),f(x'))<\varepsilon. \end{align*} Thus for every $\varepsilon>0$ there exists $\delta>0$, namely the $\delta$ defined above, such that for all $x,x'\in X$, the implication $d_X(x,x')<\delta \implies d_Y(f(x),f(x'))<\varepsilon$ holds. Therefore $f$ is uniformly continuous. [/step]