Let $(S,d)$ and $(S^*, d^*)$ be metric spaces. If $f:S \to S^*$ is uniformly continuous and if $(s_n)$ is a Cauchy sequence in $S$, then $(f(s_n))$ is a Cauchy sequence in $S^*$.
$f:S \to S^*$ is uniformly continuous: $(\forall \varepsilon >0)( \exists > 0)(\forall s,t \in S)$ $d(s,t) < \delta \implies d(f(s), f(t)) < \varepsilon$.
$(s_n)$ Cauchy: $(\forall \varepsilon > 0)(\exists N \in \mathbb{N})$ such that $\forall n,m \geq N$ we have $d(s_n,s_m)) < \varepsilon$.
Now I need to show that $(\forall \varepsilon > 0)(\exists N \in \mathbb{N})$ such that $\forall n,m \geq N$ we have $d(f(s_n),f(s_m)) < \varepsilon$.
Since $f$ is uniformly continuous, $\delta$ will depend on $\varepsilon$ only. Then I separate the problem into two cases:
Suppose $\varepsilon < \delta$: Since I know $(s_n)$ is Cauchy, I can take the same $N$ to get $d(s_n,s_m) < \varepsilon < \delta$ which implies $d(f(s_n),f(s_m)) < \varepsilon$ by uniform continuity.
Suppose $\delta \leq \varepsilon$: Since $f$ is uniformly continuous, $d(s,t) < \delta \implies d(f(s),f(t)) < \varepsilon$ will hold for all $s,t \in S$. So let $N=1$, then $d(s_n,s_m) < \delta$ which implies $d(f(s_n),f(s_m)) < \varepsilon$.
Could someone give me feedback on my proof?
Edit For every $\varepsilon > 0$, since $f$ is uniformly continuous, there exists a $\delta >0$ such that for all $s, t \in S$, $d(s, t) < \delta \implies d(f(s),f(t)) < \varepsilon$. Then for any $\delta >0$ there exists an $N \in \mathbb{N}$ such that for all $n,m \geq N$, $d(s_n, s_m) < \delta$. By uniform continuity, $d(s_n, s_m) < \delta \implies d(f(s_n),f(s_m)) < \varepsilon$.