$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\CC}{\mathcal C}\newcommand{\R}{\mathbb R}$For any real $\ep>0$, let
\begin{equation*}
Y_{n,\ep}(t):=\min(c+\ep,\max(c-\ep,Y_n(t));
\end{equation*}
here and else where, $t\in[0,T]$.
Then the process $Y_{n,\ep}$ is adapted to the filtration generated by $B$. So, we can introduce
\begin{equation*}
Z_{n,\ep}(t):=\int_0^t Y_{n,\ep}(s)\,dB(s)
\end{equation*}
and
\begin{equation*}
D_{n,\ep}(t):=Z_{n,\ep}(t)-Z(t)=\int_0^t (Y_{n,\ep}(s)-c)\,dB(s).
\end{equation*}
Then $D_{n,\ep}$ is a martingale and hence $D_{n,\ep}^2$ is a nonnegative submartingale. So, for any real $\de>0$, by the Doob martingale inequality and the Itô isometry, denoting by $\|\cdot\|$ the norm in $\CC[0,T]$, we have
\begin{equation*}
\begin{aligned}
P(\|Z_n-Z\|>\de,\|Y_n-c\|\le\ep)&\le P(\|D_{n,\ep}\|>\de) \\
&=P(\max_{t\in[0,T]}D_{n,\ep}(t)^2>\de^2) \\
&\le\frac{ED_{n,\ep}(T)^2}{\de^2} \\
&=\frac1{\de^2}\,\int_0^T E((Y_{n,\ep}(s)-c)^2\,dt \\
&\le\frac{\ep^2 T}{\de^2}\le\de
\end{aligned}
\end{equation*}
if $\ep^2\le T\de^3$.

Also, the condition $(X_n,Y_n)\Rightarrow(X,c)$ implies $Y_n\Rightarrow c$ and hence $Y_n\to c$ in probability, so that $P(\|Y_n-c\|>\ep)\to0$. So,
\begin{equation*}
\limsup_n P(\|Z_n-Z\|>\de)\le\limsup_n P(\|Z_n-Z\|>\de,\|Y_n-c\|\le\ep)+\lim_n P(\|Y_n-c\|>\ep)
\le\de+0
\end{equation*}
for any real $\de>0$.

So, $Z_n\to Z$ in probability and hence $Z_n\Rightarrow Z$. So, for any real $\de>0$ and some compact $K_{Z,\de}$ in $\CC[0,T]$ and all $n$ we have
\begin{equation*}
P(Z_n\notin K_{Z,\de})+P(Z\notin K_{Z,\de})\le\de.
\end{equation*}
Similarly, the condition $(X_n,Y_n)\Rightarrow(X,c)$ implies $X_n\Rightarrow X$ and hence for some compact $K_{X,\de}$ in $\R$ and all $n$ we have
\begin{equation*}
P(X_n\notin K_{X,\de})\le\de.
\end{equation*}

Now take any bounded (say by $1$) continuous function $f\colon\R\times\CC[0,T]$. Then $f$ is uniformly continuous on the compact set $K_{X,\de}\times K_{Z,\de}$. So, for each real $\ep>0$ there is some real $\de>0$ such that
\begin{equation*}
\begin{aligned}
&E|f(X_n,Z_n)-f(X_n,Z)|\,1(\{(X_n,Z_n),(X_n,Z)\}\subseteq K_{X,\de}\times K_{Z,\de}) \\
&\le\ep+2P(\|Z_n-Z\|>\de)\to\ep.
\end{aligned}
\end{equation*}
So,
\begin{equation}
\limsup_n E|f(X_n,Z_n)-f(X_n,Z)|\le\ep+2P(Z_n\notin K_{Z,\de})+2P(Z\notin K_{Z,\de})
+2P(X_n\notin K_{X,\de})\le\ep+4\de.
\end{equation}
Also, because $X_n\Rightarrow X$ and $X_n$ is independent of $B$ and hence of $Z$, we have $Ef(X_n,Z)\to Ef(X,Z)$.

Thus, $Ef(X_n,Z_n)\to Ef(X,Z)$. $\quad\Box$

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