[proofplan]
We prove the uniform $\varepsilon$-$\delta$ condition directly from the Lipschitz estimate. The only point requiring separation is the case $c=0$, since the usual choice $\delta=\varepsilon/c$ is then unavailable. When $c>0$, the Lipschitz inequality converts the bound $d_X(x,y)<\delta$ into the required bound on $d_Y(f(x),f(y))$; when $c=0$, the map has zero distance between all image values. The contraction case follows because a contraction is, by definition, a map satisfying such a Lipschitz estimate with contraction constant less than $1$.
[/proofplan]
[step:Choose a uniform radius for each $\varepsilon>0$]
Let $\varepsilon>0$ be arbitrary. If $c>0$, define $\delta=\varepsilon/c$; if $c=0$, define $\delta=1$. In either case, $\delta>0$.
[guided]
To prove [uniform continuity](/page/Uniform%20Continuity), we must show that for every tolerance $\varepsilon>0$ in the codomain metric $d_Y$, there is a single radius $\delta>0$ in the domain metric $d_X$ that works for all pairs of points $x,y \in X$. Fix such an arbitrary $\varepsilon>0$.
The Lipschitz estimate in the hypothesis says that distances after applying $f$ are bounded by $c$ times the original distance. If $c>0$, the natural way to make $c\,d_X(x,y)$ smaller than $\varepsilon$ is to force $d_X(x,y)<\varepsilon/c$. Thus in this case we set
\begin{align*}
\delta=\varepsilon/c.
\end{align*}
This is positive because both $\varepsilon>0$ and $c>0$.
If $c=0$, division by $c$ is not available. In that case the Lipschitz estimate is even stronger: it will force all image distances to be zero. Therefore any positive radius is sufficient, and we choose
\begin{align*}
\delta=1.
\end{align*}
This gives a positive number $\delta$ in every case.
[/guided]
[/step]
[step:Verify the uniform continuity condition using the Lipschitz bound]
Let $x,y \in X$ satisfy $d_X(x,y)<\delta$.
If $c>0$, then by the assumed Lipschitz estimate and the definition of $\delta$,
\begin{align*}
d_Y(f(x),f(y)) \le c\,d_X(x,y) < c\delta = \varepsilon.
\end{align*}
If $c=0$, then the Lipschitz estimate gives
\begin{align*}
d_Y(f(x),f(y)) \le 0\cdot d_X(x,y)=0.
\end{align*}
Since $d_Y$ is a metric, $d_Y(f(x),f(y))\ge 0$, hence
\begin{align*}
d_Y(f(x),f(y))=0<\varepsilon.
\end{align*}
Thus for every $\varepsilon>0$ there exists $\delta>0$ such that, for all $x,y \in X$,
\begin{align*}
d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon.
\end{align*}
This is exactly uniform continuity of $f:X\to Y$.
[guided]
Now take arbitrary points $x,y \in X$ satisfying
\begin{align*}
d_X(x,y)<\delta.
\end{align*}
We must prove
\begin{align*}
d_Y(f(x),f(y))<\varepsilon.
\end{align*}
First suppose $c>0$. The hypothesis applies to this particular pair $x,y \in X$, so
\begin{align*}
d_Y(f(x),f(y)) \le c\,d_X(x,y).
\end{align*}
Because $d_X(x,y)<\delta$ and $c>0$, multiplying the strict inequality by $c$ preserves its direction:
\begin{align*}
c\,d_X(x,y)<c\delta.
\end{align*}
By the definition $\delta=\varepsilon/c$, we have
\begin{align*}
c\delta=c(\varepsilon/c)=\varepsilon.
\end{align*}
Combining these inequalities gives
\begin{align*}
d_Y(f(x),f(y))<\varepsilon.
\end{align*}
Now suppose $c=0$. The same Lipschitz estimate gives
\begin{align*}
d_Y(f(x),f(y)) \le 0\cdot d_X(x,y)=0.
\end{align*}
Every metric takes nonnegative values, so
\begin{align*}
d_Y(f(x),f(y))\ge 0.
\end{align*}
Therefore
\begin{align*}
d_Y(f(x),f(y))=0.
\end{align*}
Since $\varepsilon>0$, this implies
\begin{align*}
d_Y(f(x),f(y))<\varepsilon.
\end{align*}
In both cases, the same $\delta$ works for all choices of $x,y \in X$ satisfying $d_X(x,y)<\delta$. This proves the uniform continuity condition:
\begin{align*}
\forall \varepsilon>0\ \exists \delta>0\ \forall x,y\in X,\quad d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon.
\end{align*}
Hence $f:X\to Y$ is uniformly continuous.
[/guided]
[/step]
[step:Apply the result to contraction mappings]
Let $(X,d_X)$ be a [metric space](/page/Metric%20Space) and let $f:X\to X$ be a [contraction mapping](/page/Contraction%20Mapping). By the definition of a contraction mapping, there exists a constant $c\in[0,1)$ such that, for all $x,y\in X$,
\begin{align*}
d_X(f(x),f(y))\le c\,d_X(x,y).
\end{align*}
Since $c\ge 0$, the first part of the theorem applies with $Y=X$ and $d_Y=d_X$. Therefore $f$ is uniformly continuous.
[guided]
Let $(X,d_X)$ be a metric space and let $f:X\to X$ be a contraction mapping. The definition of contraction mapping supplies a constant $c\in[0,1)$ such that every pair $x,y\in X$ satisfies
\begin{align*}
d_X(f(x),f(y))\le c\,d_X(x,y).
\end{align*}
This is exactly the Lipschitz hypothesis from the first part of the theorem, with the codomain metric equal to the domain metric: take $Y=X$ and $d_Y=d_X$. The interval condition $c\in[0,1)$ also gives $c\ge 0$, so all hypotheses of the first part are satisfied. Therefore the first part implies that $f:X\to X$ is uniformly continuous.
[/guided]
[/step]
