= Solution
For a <metric space> $(X,d)$, <uniform continuity> means
$$
\forall\varepsilon>0\ \exists\delta>0\ \forall x,y\in X:
\quad d(x,y)<\delta\ \Longrightarrow\ |f(x)-f(y)|<\varepsilon.
$$
The choice of $\delta$ is independent of the points. A <Lipschitz continuous> function satisfies $|f(x)-f(y)|\leq Ld(x,y)$ for one finite constant $L\geq0$. Choosing $\delta=\varepsilon/(L+1)$ proves \b[Lipschitz continuity implies uniform continuity].
Let $(x_n)$ be a <Cauchy sequence>. For a given $\varepsilon>0$, uniform continuity supplies $\delta>0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\geq N$. Then $|f(x_m)-f(x_n)|<\varepsilon$, so $(f(x_n))$ is Cauchy in $\mathbb R$. By <completeness of the real numbers>, \b[$(f(x_n))$ converges], even if the original sequence has no limit in $X$.
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