Having read Mateusz Kwaśnicki's answer, I will now write it in my own way:
Lemma. Let $S_\infty$ and $T$ be separable metric spaces, and let $(S_j)_{j \in \mathbb{N}}$ be a sequence of Borel subsets of $S_\infty$. Let $\{F_j\}_{j \in \mathbb{N} \cup \{\infty\}}$ be a family of Borel-measurable functions $F_j \colon S_\infty \to T$, with $F_\infty$ being continuous. Suppose that for every $(x_j)_{j \in \mathbb{N} \cup \{\infty\}} \in \prod_{j \in \mathbb{N} \cup \{\infty\}} S_j$, if $x_j \to x_\infty$ as $j \to \infty$ then $F_j(x_j) \to F_\infty(x_\infty)$ as $j \to \infty$. Then for any family $(\Xi_j)_{j \in \mathbb{N} \cup \{\infty\}}$ of $S_\infty$-valued random variables $\Xi_j$ with $\Xi_j \in S_j$ almost surely, if $\Xi_j$ converges in law to $\Xi_\infty$ then $F_j(\Xi_j)$ converges in law to $F_\infty(\Xi_\infty)$.
Proof of the result given the Lemma.
Let $M=\mathbb{E}[\Delta_i]=\mathbb{E}[X_i^2]$. Let $S_\infty=C([0,\infty),\mathbb{R}^2)$, and for each $\lambda \in (0,\infty)$ let
$$ S_\lambda \ = \ \left\{ t \mapsto \begin{pmatrix} \sqrt{\lambda}(w(t)-Mt) \\ v(t) \end{pmatrix} \, : \, w \in \mathrm{Homeo}([0,\infty)), v \in C([0,\infty),\mathbb{R}) \right\}. $$
For each $\lambda \in (0,\infty)$ and $n \in \mathbb{N}$, let
$$ \Xi_\lambda(\tfrac{n}{\lambda}) \ = \ \begin{pmatrix} \frac{1}{\sqrt{\lambda}} \sum_{i=1}^n (\Delta_i - M) \\ \frac{1}{\sqrt{M\lambda}} \sum_{i=1}^n X_i \end{pmatrix} $$
and extend by linear interpolation: for $r \in (0,1)$, writing $t=\frac{n+r}{\lambda}$, we define
\begin{align*}
\Xi_\lambda(t) \ &= \ r\Xi_\lambda(\tfrac{n+1}{\lambda})+(1-r)\Xi_\lambda(\tfrac{n}{\lambda}) \\
&= \begin{pmatrix} \left( \frac{1}{\sqrt{\lambda}} \left( r\Delta_{n+1} + \sum_{i=1}^n \Delta_i \right) \right) - \sqrt{\lambda}Mt \\ \frac{1}{\sqrt{M\lambda}} \left( rX_{n+1} + \sum_{i=1}^n X_i \right). \end{pmatrix}
\end{align*}
Hence, for each $\lambda \in (0,\infty)$, we have that $\Xi_\lambda\overset{\mathrm{def}}{=}(\Xi_\lambda(t))_{t \geq 0}$ belongs almost surely to $S_\lambda$, with the corresponding $w \in \mathrm{Homeo}([0,\infty))$ given by
$$ w(\tfrac{n}{\lambda}) \ = \ \frac{1}{\lambda} \sum_{i=1}^n \Delta_i . $$
The Donsker invariance principle gives that as an $S_\infty$-valued random variable, $\Xi_\lambda$ is convergent in distribution as $\lambda \to \infty$, to a two-dimensional Brownian motion $\Xi_\infty\!=\!(\Xi_\infty(t))_{t \geq 0}$ whose second-coordinate projection is a one-dimensional Wiener process (i.e. standard Brownian motion).
Now for each $\lambda \in (0,\infty)$, define $F_\lambda \colon S_\lambda \to C([0,\infty),\mathbb{R})$ by
$$ F_\lambda\!\begin{pmatrix} u \\ v \end{pmatrix} \ = \ \sqrt{M}v \circ \left( M.\mathrm{id}_{[0,\infty)} + \tfrac{1}{\sqrt{\lambda}} u \right)^{\!-1}, $$
and define $F_\infty \colon S_\infty \to C([0,\infty),\mathbb{R})$ by
$$ F_\infty\!\begin{pmatrix} u \\ v \end{pmatrix}(t) \ = \ \sqrt{M}v\!\left( \tfrac{1}{M} t \right). $$
So $F_\lambda(\Xi_\lambda) = (W_t^{(\lambda)})_{t \geq 0}$ for each $\lambda \in (0,\infty)$, and $F_\infty(\Xi_\infty)$ is a one-dimensional Wiener process.
Now for any $\lambda_n \nearrow \infty$, any $(x_n)_{n \in \mathbb{N}}=\!\left( \begin{pmatrix} u_n \\ v_n \end{pmatrix} \right)_{\!\!n \in \mathbb{N}} \in \prod_{n \in \mathbb{N}} S_{\lambda_n}$ and any $x_\infty=\!\begin{pmatrix} u_\infty \\ v_\infty \end{pmatrix} \in S_\infty$, if $x_n$ converges uniformly on bounded sets to $x_\infty$, then $M.\mathrm{id}_{[0,\infty)} + \frac{1}{\sqrt{\lambda_n}} u_n$ converges uniformly on bounded sets to $M.\mathrm{id}_{[0,\infty)}$, so $\left( M.\mathrm{id}_{[0,\infty)} + \frac{1}{\sqrt{\lambda_n}} u_n \right)^{\!-1}$ converges uniformly on bounded sets to $\frac{1}{M}\mathrm{id}_{[0,\infty)}$; and also $v_n$ converges uniformly on bounded sets to $v_\infty$; and therefore $F_{\lambda_n}(x_n)$ converges uniformly on bounded sets to $F_\infty(x_\infty)$.
Hence the Lemma gives the desired result.
Proof of the Lemma.
For convenience of notation, assume that all the random variables are over a probability space $(\Omega,\mathcal{F},\mathbb{P})$.
Let $d$ be the metric on $T$. For each $\varepsilon>0$, define the decreasing sequence $(G_j(\varepsilon))_{j \in \mathbb{N}}$ of subsets of $S_\infty$ by
$$ G_j(\varepsilon) \ = \ \overline{\bigcup_{i=j}^\infty \{x \in S_i : d(F_i(x),F_\infty(x)) \geq \varepsilon \}}. $$
We first show that for all $\varepsilon>0$, $\bigcap_{j \in \mathbb{N}} G_j(\varepsilon)=\emptyset$. If we have $x_\infty \in \bigcap_{j \in \mathbb{N}} G_j(\varepsilon)$, we can construct a sequence $n_j \nearrow \infty$ and a sequence $x_{n_j} \in S_{n_j}$ converging to $x_\infty$ such that $d(F_{n_j}(x_{n_j}),F_\infty(x_{n_j})) \geq \varepsilon$ for all $j \in \mathbb{N}$; and since $F_\infty(x_{n_j}) \to F_\infty(x_\infty)$ (by continuity of $F_\infty$), it follows that $F_{n_j}(x_{n_j})$ does not converge to $F_\infty(x_\infty)$, contradicting our assumption.
Now fix any family $(\Xi_j)_{j \in \mathbb{N} \cup \{\infty\}}$ of $S_\infty$-valued random variables $\Xi_j$ with $\Xi_j \in S_j$ almost surely, such that $\Xi_j$ converges in law to $\Xi_\infty$. For each $\varepsilon>0$ we have that $\mathbb{P}(\Xi_j \in G_j(\varepsilon)) \to 0$, since
\begin{align*}
\limsup_{j \to \infty} \mathbb{P}(\Xi_j \in G_j) \ &\leq \ \lim_{m \to \infty} \limsup_{j \to \infty} \mathbb{P}(\Xi_j \in G_m) \\
&\leq \ \lim_{m \to \infty} \mathbb{P}(\Xi_\infty \in G_m) \quad \text{since $\Xi_j$ converges in law to $\Xi_\infty$} \\
&= \ 0 \quad \text{since } \bigcap_{m \in \mathbb{N}} G_m=\emptyset.
\end{align*}
Now to show the desired convergence, fix a bounded Lipschitz function $g$ on $T$ which, without loss of generality, we assume to map into $[0,1]$ and to have Lipschitz constant $1$. Fix $\varepsilon>0$. Let $N$ be sufficiently large that for all integers $j \geq N$,
$$ \mathbb{P}(\Xi_j \in G_j(\tfrac{\varepsilon}{3})) \ < \ \tfrac{\varepsilon}{3} $$
and
$$ \big| \mathbb{E}[g(F_\infty(X_j))] - \mathbb{E}[g(F_\infty(X_\infty))] \big| < \tfrac{\varepsilon}{3}, $$
where the latter is possible by the convergence in law of $X_j$ to $X_\infty$, since $g \circ F_\infty$ is a bounded continuous function on $S_\infty$. Using the former of these two statements, an easy calculation yields
$$ \big| \mathbb{E}[g(F_j(X_j)) - g(F_\infty(X_j))] \big| < \tfrac{2\varepsilon}{3}, $$
and combining this with the latter gives
$$ \big| \mathbb{E}[g(F_j(X_j))] - \mathbb{E}[g(F_\infty(X_\infty))] \big| < \varepsilon $$
as required.
