Show directly that if $\{s_n\}$ is a Cauchy sequence then so is $\{|s_n|\}$. From this conclude that $\{|s_n|\}$ converges whenever $\{s_n\}$ converges.
Let $\{s_n\}$ be a Cauchy sequence. Then by definition, for any given $\varepsilon>0$ there exists $m>0$ such that $|s_n-s_m|<\varepsilon$ for all $n\geq m$. Then we have $$||s_n|-|s_m||\leq|s_n-s_m|$$ Therefore, from the definition $$||s_n|-|s_m||\leq|s_n-s_m|<\varepsilon$$ for all $n\geq m$. Hence, $\{|s_n|\}$ is a Cauchy sequence.
And then to prove that convergence of $\{s_n\}$ implies the convergence of $\{|s_n|\}$:
Let $\varepsilon>0$. If $\{s_n\}$ converges to $L$, then there exists $N$ such that $|s_n-L|<\varepsilon$, whenever $n\geq N$. Hence, for $n\geq N$, we have $||s_n|-L|\leq |s_n-L|<\varepsilon$. Thus $\{|s_n|\}$ converges to $|L|$.
That's how I proved but I'm not sure if I possibly made some mistakes or missed some steps!?
If we are dealing with a complete metric space, we know that every Cauchy sequence is convergent to a limit in that metric space. So if you prove that the sequence is $\{|s_n| \}$ is Cauchy, indeed by using the reverse triangle inequality, we automatically get that it is a convergent sequence.