Proof of A Lipschitz Map is Uniformly Continuous

Let $\Lambda$ be a nonnegative real number such that

$$d_Y\bigl(f(x),f(x')\bigr)\le\Lambda\,d_X(x,x')\qquad\text{for all }x,x'\in X,$$

as provided by Lipschitz Map Between Metric Spaces.

Uniform continuity. Let $\varepsilon$ be a real number with $0<\varepsilon$. Since $0\le\Lambda$ we have $0<1\le\Lambda+1$, so

$$\delta=\frac{\varepsilon}{\Lambda+1}$$

is a real number with $0<\delta$. Let $x,x'\in X$ satisfy $d_X(x,x')<\delta$. Multiplying this inequality by the nonnegative number $\Lambda$ preserves the order, so $\Lambda\,d_X(x,x')\le\Lambda\delta$. Moreover $\Lambda<\Lambda+1$ and $0<\varepsilon$ give $\Lambda\varepsilon<(\Lambda+1)\varepsilon$, and dividing by the positive number $\Lambda+1$ preserves the strict inequality, so

$$\Lambda\delta=\frac{\Lambda\varepsilon}{\Lambda+1}<\varepsilon .$$

Combining the three estimates,

$$d_Y\bigl(f(x),f(x')\bigr)\le\Lambda\,d_X(x,x')\le\Lambda\delta<\varepsilon .$$

Since $\varepsilon$ was arbitrary and $\delta$ was chosen independently of $x$ and $x'$, the map $f$ is uniformly continuous on $X$ in the sense of Uniformly Continuous Map Between Metric Spaces.

Continuity. Let $x\in X$ and let $\varepsilon$ be a real number with $0<\varepsilon$, and let $\delta$ be as above. Every $y\in X$ with $d_X(x,y)<\delta$ satisfies, by the displayed estimate applied to the pair $x,y$,

$$d_Y\bigl(f(y),f(x)\bigr)=d_Y\bigl(f(x),f(y)\bigr)<\varepsilon ,$$

the first equality being the symmetry of $d_Y$ (condition 3 of Metric Space). Hence $f$ is continuous at $x$ relative to $X$ in the sense of Continuous Map Between Metric Spaces. As $x\in X$ was arbitrary, $f$ is continuous on $X$.