This one is a simple case of combininingcombining two limits into one:. In the present case the condition is necessary and sufficient, so you have a stronger result than the one you are positing.
Theorem: If $X_n\overset{d}{\to}X$ andSuppose that $\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$ then $Y_n\overset{d}{\to}X$. Then for any random variable $L$ we have: $$X_n\overset{d}{\to}L \quad \iff \quad Y_n\overset{d}{\to}L.$$
Proof: Let $f$ be an arbitrary bounded continuous function. Since $X_n\overset{d}{\to}X$ and $\mathcal{L}(Y_n) \overset{w.a.}{\longleftrightarrow}\mathcal{L}(X_n)$, so we have the respective limits: $$\begin{aligned} &\lim_{n \to \infty} \mathbb{E}(f(X_n)) - \mathbb{E}(f(X)) = 0, \\[6pt] &\lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) = 0. \\[6pt] \end{aligned}$$ We therefore obtain the limit of interest: $$\lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) = 0.$$ $$\begin{aligned} \text{Limit} &\equiv \lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X)) \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) + \mathbb{E}(f(X_n)) - \mathbb{E}(f(X)) \Big] \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) \Big] + \lim_{n \to \infty} \Big[ \mathbb{E}(f(X_n)) - \mathbb{E}(f(X)) \Big] \\[6pt] &= 0+0 = 0. \\[6pt] \end{aligned}$$($\implies$) This establishes thatSince $Y_n\overset{d}{\to}X$,$X_n\overset{d}{\to}L$ we have $\lim_{n \to \infty} \mathbb{E}(f(X_n)) - \mathbb{E}(f(L)) = 0$ which was to be showngives: $$\begin{aligned} \text{Limit} &\equiv \lim_{n \to \infty} \mathbb{E}(f(Y_n)) - \mathbb{E}(f(L)) \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) + \mathbb{E}(f(X_n)) - \mathbb{E}(f(L)) \Big] \\[6pt] &= \lim_{n \to \infty} \Big[ \mathbb{E}(f(Y_n)) - \mathbb{E}(f(X_n)) \Big] + \lim_{n \to \infty} \Big[ \mathbb{E}(f(X_n)) - \mathbb{E}(f(L)) \Big] \\[6pt] &= 0+0 = 0. \\[6pt] \end{aligned}$$ The proof of the reverse implication ($\impliedby$) is identical. $\blacksquare$