ADD You got to
$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }}$
Since $s_n\to s$, for every $\epsilon >0$ there is an $n_0$ for wich $\left| {{s_n} - s} \right| < \varepsilon \sqrt s $ whenever $n\geq n_0$ (i.e. $\varepsilon \sqrt s$ is also an $\epsilon'>0$). Then, for this $n_0$, $\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{{\left| {{s_n} - s} \right|}}{{\sqrt s }} < \frac{{\varepsilon \sqrt s }}{{\sqrt s }} = \varepsilon $
which means $\sqrt {s_n}\to\sqrt s$.
You're almost done. You arrived at
$\left|\dfrac{s_n - s}{\sqrt{s_n}+\sqrt{s}}\right| $
You know that $s_n\to s$, so you can make $|s_n-s|$ as small as you wish. Now, we need to know how to handle $\sqrt{s_n}+\sqrt{s}$. Since $s_n\to s$, there is an $n_0$ for wich
$|s-s_n|<3s/4$
Since $s_n>0$,$s\geq 0$, then
$s-s_n\leq|s-s_n|<3s/4$
This means that
$s_n>s/4$
then
$2\sqrt {{s_n}} > \sqrt s $
or, since $\sqrt s>0$
$\eqalign{ & 2\sqrt {{s_n}} + 2\sqrt s > 3\sqrt s \cr & \sqrt {{s_n}} + \sqrt s > \frac{{3\sqrt s }}{2} \cr & \frac{1}{{\sqrt {{s_n}} + \sqrt s }} < \frac{2}{{3\sqrt s }} \cr} $
Again, since $s_n\to s$, there is an $n_1$ for wich
$|s-s_n|<\epsilon \frac{3\sqrt s}{{2 }}$
Then, taking $n\geq \max\{n_0,n_1\}$ we have
$\left| {\frac{{{s_n} - s}}{{\sqrt {{s_n}} + \sqrt s }}} \right| < \frac{2}{{3\sqrt s }}\left| {{s_n} - s} \right| < \frac{2}{{3\sqrt s }}\frac{{3\sqrt s }}{2}\varepsilon = \varepsilon $