Not only is $S$ closed, it’s also bounded and therefore compact, so it’s certainly complete.
Correction: I somehow misread the definition of $W$ as $\Bbb Z^+\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, so what I originally said about it is nonsense.
$W=\left\{n+\frac1n:n\in\Bbb Z^+\right\}=\left\{2,\frac52,\frac{10}3,\dots,\frac{n^2+1}n\dots\right\}\;,$
which is a closed, discrete set and therefore complete. The minimum distance between any two distinct points of $W$ is $1/2$, between $2$ and $5/2$. To prove this, suppose that $m,n\in\Bbb Z^+$ with $m. Then since $\dfrac1m\le 1$, $m+\dfrac1m\le n, and
$\begin{align*} \left|\left(m+\frac1m\right)-\left(n+\frac1n\right)\right|&=n+\frac1n-\left(m+\frac1m\right)\\ &=n-m-\left(\frac1m-\frac1n\right)\\ &=n-m-\frac{n-m}{mn}\;. \end{align*}$
Now $n-m, and $mn\ge n$, so $\dfrac{n-m}{mn}<1$, and $n-m-\dfrac{n-m}{mn}>n-m-1$. This is at least $1$ unless $n=m+1$.
If $n=m+1$, $n-m-\dfrac{n-m}{mn}=1-\dfrac1{m(m+1)}$ is maximized when $m=1$, in which case it’s $\dfrac12$.
In all cases, therefore,
$\left|\left(m+\frac1m\right)-\left(n+\frac1n\right)\right|\ge\frac12$ when $1\le m.