Does Skorohod Dudley Wichura theorem imply Continuous Mapping Theorem?

Question

The continuous mapping theorem (CMT) asserts that for $g$ a.e. continuous on the support of $X$,

1. $X_n\overset{a.s.}{\rightarrow}X\implies g(X_n)\overset{a.s.}{\rightarrow}g(X).$
2. $X_n\overset{p}{\rightarrow}X\implies g(X_n)\overset{p}{\rightarrow}g(X).$
3. $X_n\overset{d}{\rightarrow}X\implies g(X_n)\overset{d}{\rightarrow}g(X).$

Skorohod Dudley Wichura (SDW) theorem, as I understand, says that if $X_n\overset{d}{\rightarrow}X$, then there is a probability space where $Y_n,Y$ are defined such that $Y_n\overset{d}{=}X_n,Y\overset{d}{=}X,Y_n\overset{a.s.}{\rightarrow}Y.$

Suppose we know 1. to hold. Can SDW be used to prove 2. of CMT?

I am able to show SDW proves 3. of CMT as follows: \begin{align*} X_n\overset{d}{\rightarrow}X&\implies \exists Y_n\overset{d}{=}X_n,Y\overset{d}{=}X,\text{ s.t. } Y_n\overset{a.s.}{\rightarrow}Y\\ &\implies g(Y_n)\overset{a.s.}{\rightarrow}g(Y)\quad \text{by 1.}\\ &\implies \underbrace{g(Y_n)}_{\overset{d}{=}g(X_n)} \overset{d}{\rightarrow} \underbrace{g(Y)}_{\overset{d}{=}g(X)}\quad \text{since } \overset{a.s.}{\rightarrow} \implies \overset{d}{\rightarrow}\\ &\implies g(X_n)\overset{d}{\rightarrow}g(X). \end{align*}

However, I am not sure how to extend this to use SDW to prove 2. of CMT since $\underbrace{g(Y_n)}_{\overset{d}{=}g(X_n)} \overset{p}{\rightarrow} \underbrace{g(Y)}_{\overset{d}{=}g(X)}\not \Rightarrow g(X_n)\overset{p}{\rightarrow}g(X).$