Statistical closeness implies computational indistinguishability This is so trivial that authors usually don't bother to give an explicit proof. But for me there is some vagueness.

We say that two ensembles $X_n$ and $Y_n$ are statistically close, if 
$$ \Delta(n) = 1/2 \sum_{\alpha}|\mathbb{P}[X_n = \alpha] - \mathbb{P}[Y_n = \alpha]| $$ is negligible in n. The probability is taken over the randomness of $X_n$ and $Y_n$ respectively.

We say that two ensembles are computationally indistinguishable if for every PPT-adversary D we have
$$ |\mathbb{P}[D(X_n) \to 1] - \mathbb{P}[D(Y_n) \to 1]|$$
is negligible.

Why the former implies the latter?

I do understand that for every deterministic function $f$ we have
$\Delta(f(X), f(Y)) \le \Delta(X, Y),$ where $\Delta(\cdot, \cdot)$ is the statistical distance. 

But in the case of PPT adversaries $D$ is not deterministic, there are implicit random coins. Why can we treat PPT-algorithm $D$ as deterministic function?
A probabilistic distinguisher is still a deterministic function of its input and random coins. So a probabilistic distinguisher trying to distinguish $X$ from $Y$ is equivalent to a deterministic distinguisher trying to distinguish $(X,R)$ from $(Y,R)$ where $R$ is a uniform distribution over random coins (importantly: independent of $X$/$Y$).

But:

\begin{align}
  \Delta\Bigl( (X,R), (Y,R) \Bigr) &amp;= \frac12 \sum_{\alpha,r} \Bigl| \Pr[(X,R)=(\alpha,r)] - \Pr[(Y,R)=(\alpha,r)]\Bigr|
\\
   &amp;= \frac12 \sum_{\alpha,r} \Bigl| \Pr[X=\alpha]\Pr[R=r] - \Pr[Y=\alpha]\Pr[R=r]\Bigr| \\
   &amp;= \frac12 \sum_{\alpha,r} \Bigl| \Pr[X=\alpha] - \Pr[Y=\alpha]\Bigr| \Pr[R=r] 
\\
   &amp;= \frac12 \sum_{\alpha} \Bigl| \Pr[X=\alpha] - \Pr[Y=\alpha]\Bigr| \;\underbrace{\sum_r \Pr[R=r]}_{=1} \\
   &amp;= \frac12 \sum_{\alpha} \Bigl| \Pr[X=\alpha] - \Pr[Y=\alpha]\Bigr|  \\
   &amp;= \Delta(X,Y)
\end{align}

In short, having access to some distribution that is independent of $X$/$Y$ doesn't help (or hurt) to distinguish $X$ from $Y$.