= Solution
Write the constant limit as $c$. First, <convergence in distribution to a constant implies convergence in probability>: the <Portmanteau theorem> applied to the closed set $\{y:|y-c|\geq\varepsilon\}$ gives
$$
\limsup_{n\to\infty}\mathbb P(|Y_n-c|\geq\varepsilon)\leq0.
$$
Now compare <characteristic functions> without imposing <independence>. For fixed $t\in\mathbb R$,
$$
\begin{aligned}
\left|\mathbb Ee^{it(X_n+Y_n)}-e^{itc}\mathbb Ee^{itX_n}\right|
&\leq\mathbb E|e^{itY_n}-e^{itc}|\\
&\leq |t|\varepsilon+2\mathbb P(|Y_n-c|>\varepsilon).
\end{aligned}
$$
The second bound uses $|e^{iu}-e^{iv}|\leq\min\{2,|u-v|\}$. By <convergence in probability>, the limiting upper bound is $|t|\varepsilon$; then let $\varepsilon\downarrow0$. Meanwhile $X_n$ has <weak convergence of random variables> to $X$, so $\mathbb Ee^{itX_n}\to\mathbb Ee^{itX}$. Thus the <characteristic function> of the sum tends to $e^{itc}\mathbb Ee^{itX}$, the <characteristic function> of $X+c$, which is <continuous> at zero. The <Lévy continuity theorem> proves
$$
\boxed{X_n+Y_n\ \Longrightarrow\ X+c.}
$$
This is the additive case of the <Slutsky theorem>, with its proof supplied. The constant limiting marginal removes the dependence obstruction in part (b).
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