Link between convergence in distribution and almost sure convergence Let $X_n$, $Y_n$ and $Y$ be random vectors. If $X_n \xrightarrow[]{d} X$ and $X_n - Y_n \xrightarrow[]{a.s.} 0$. Can we prove that $Y_n \xrightarrow[]{d} X$ ?

Yes, we can.
First, define $Z_n := X_n-Y_n$
Now,
\begin{align} 
Z_n \overset{a.s}{\to} 0   & \implies Z_n \overset{p}{\to} 0 \quad \text{(van der Vaart, Thm. 2.7. (i))} \\
& \iff \underset{n \to \infty}{\lim}\mathbb{P}(|Z_n-0|>\epsilon)=0, \forall \epsilon>0 \quad \text{(definition of $\overset{p}{\to}$)} \\
& \iff \underset{n \to \infty}{\lim}\mathbb{P}(|(-Z_n)-0|>\epsilon)=0, \forall \epsilon>0 \quad \text{(as $|Z_n-0|=|Z_n|=|-Z_n|=|(-Z_n)-0|$)} \\
& \iff (-Z_n) \overset{p}{\to} 0
\end{align}
where the reference is: Vaart, A. (1998). Asymptotic Statistics (Cambridge Series in Statistical and Probabilistic Mathematics). Cambridge: Cambridge University Press.
Further, we have $Y_n = X_n - (X_n-Y_n) = X_n - Z_n = X_n + (-Z_n)$
So, as $X_n \overset{d}{\to} X$ and $(-Z_n) \overset{p}{\to} 0 $, we have by Slutsky's Lemma (van der Vaart Lemma 2.8 (i)) that
$Y_n = X_n + (-Z_n) \overset{d}{\to} X + 0 = X$ i.e. $Y_n \overset{d}{\to} X$
