This proof is of course incorrect. In particular, the inequality $\Pr[X_i \neq Y_i] \leq \Pr[X_i^\ast \neq Y_i^\ast]$ actually goes in the opposite direction.

A correct proof is as follows. For each $i$, let $(X_i^*,Y_i^*)$ be the closest coupling of $(X_i,Y_i)$, so that $d(X_i,Y_i)=P(X_i^*\ne Y_i^*)$. Suppose also that the random pairs $(X_i^*,Y_i^*)$ are independent. Let $S_n^*:=\sum_{i=1}^n X_i^*$ and $T_n^*:=\sum_{i=1}^n Y_i^*$. Then $S_n^*$ equals $S_n$ in distribution and $T_n^*$ equals $T_n$ in distribution (because the $X_i$'s are independent and the $Y_i$'s are independent). So, \begin{equation} d(S_n,T_n)=d(S_n^*,T_n^*)\le P(S_n^*\ne T_n^*)\le\sum_{i=1}^n P(X_i^*\ne Y_i^*) =\sum_{i=1}^n d(X_i,Y_i). \end{equation}

This proof is of course incorrect. In particular, the inequality $\Pr[X_i \neq Y_i] \leq \Pr[X_i^\ast \neq Y_i^\ast]$ actually goes in the opposite direction (this inequality does not seem to be in the paper linked in your post).

A correct proof is as follows. For each $i$, let $(X_i^*,Y_i^*)$ be the closest coupling of $(X_i,Y_i)$, so that $d(X_i,Y_i)=P(X_i^*\ne Y_i^*)$. Suppose also that the random pairs $(X_i^*,Y_i^*)$ are independent. Let $S_n^*:=\sum_{i=1}^n X_i^*$ and $T_n^*:=\sum_{i=1}^n Y_i^*$. Then $S_n^*$ equals $S_n$ in distribution and $T_n^*$ equals $T_n$ in distribution (because the $X_i$'s are independent and the $Y_i$'s are independent). So, \begin{equation} d(S_n,T_n)=d(S_n^*,T_n^*)\le P(S_n^*\ne T_n^*)\le\sum_{i=1}^n P(X_i^*\ne Y_i^*) =\sum_{i=1}^n d(X_i,Y_i). \end{equation}