 Solution

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.