Below is Theorem 3.9 and its proof from Billingsley's Convergence in Probability Measures. I find this proof quite odd and I would greatly appreciate some help understanding the proof. So the proof is complete once we establish (3.38) by a previous theorem.
To verify (3.38) for $X'' \equiv a''$ the author sets $a'' \notin \partial A''$. Why is this so? Shouldn't we have $a'' \in \partial A''$ since $A''$ is a $a''$-continuity set by assumption, which means $P[a'' \in \partial A'']=0$? And why does this lead to $P[X_n^{''}\notin A''] \to 0$ if $a'' \in A''$? Also I cannot figure out how this combined with $X_n' \Rightarrow X'$ gives (3.38)? The inequalities below give me even less clue as to how they are proceeding with the proof. Finally, if $a'' \notin A''$, why is $P[X_n^{''} \in A'']\to 0$ and how does this give (3.38)?
This part is very confusing to me. I would greatly appreciate some help.
I think you got the definition of a continuity set backwards. A set $A$ is a continuity set with reference to some random variable $X$, if $\mathrm{P}(X\in\partial A)=0$, i.e. the boundary is negligible (in a probabilistic sense).
Since the weak limit $a''$ is here nonstochastic, the provision that $A''$ is a continuity set is $\mathrm{P}(a''\in \partial A'')=\mathbf{1}(a''\in \partial A'')=0$, i.e. $a''\notin \partial A''$. If $a''$ belonged to $\partial A''$, the entire "distribution" of $a''$ would sit on the boundary, and it would certainly not be negligible.
Since $A''$ is an $a''$-continuity set, Portmanteau theorem gives us $\mathrm{P}(X''_n\in A'')\to \mathrm{P}(a''\in A'')=\mathbf{1}(a''\in A'')$. Billingsley next splits the proof into the two cases: (i) $a''\in A''$ and (ii) $a''\notin A''$.
In Case (i), $\mathrm{P}(X''_n\in A'')\to \mathbf{1}(a''\in A'')=1$. Now, use that for sets $A$ and $B$, we can always contain $A\subseteq(A\cap B)\cup B^c$. Probability measures are subadditive, so rewriting quantities in terms of events involving random variables and rearranging gives you the very first inequality. The second follows from containment. The lower and upper bounds both have the same limit (Portmanteau again), which is $\mathrm{P}(X' \in A')$. The squeeze theorem therefore tells us that $\mathrm{P}(X'_n\in A', X''_n\in A'')\to \mathrm{P}(X' \in A')$. The right-hand side is exactly $\mathrm{P}(X' \in A')=\mathrm{P}(X' \in A')\mathbf{1}(a''\in A'')=\mathrm{P}(X' \in A',a''\in A'')$, using our case and independence/constancy.
In Case (ii), $\mathrm{P}(X''_n\in A'')\to \mathbf{1}(a''\in A'')=0$. The stated inequality follows from containment, and zero is exactly $\mathrm{P}(X'\in A', a''\in A'')$, again using our case and independence/constancy. So in either case, $\mathrm{P}(X'_n\in A', X''_n\in A'')$ has the desired limit.