Is the Poincaré dual of the first Stiefel-Whitney class of a manifold necessarily orientable? I think the answer should be yes, and I think there's an argument for it for triangulable (compact) manifolds as follows:

For our $n$-manifold $M$, given a triangulation pick some $(n-1)$-simplices that give a submanifold (denoted $N$) dual to $[w_1(TM)]$. Then consider the barycentric subdivision of the entire triangulation. Away from the dual $N$, it will be possible to consistently label the $n$-simplices of $M$ as $+$ or $-$ depending on the orientation of the simplex. Since we have a barycentric subdivided triangulation, away from $N$ this labeling will label neighboring simplices by opposite signs. The border of where this opposite labeling fails is given by a set of $(n-1)$-simplices consisting of precisely $N$.
So, $N$ can be thought of as consisting of the boundaries of $n$-simplices that both share the same $\pm$ sign. But, this means that the $(n-1)$-simplices that comprise $N$ can be given a consistent labeling of $+/-$ signs that come from the shared sign of the two $n$-simplices that share an $(n-1)$-simplex of $N$. These signs will alternate between neighboring $(n-1)$-simplices of $N$, and since we are considering a barycentric subdivided triangulation of $N$, this shows that $N$ is orientable

I'm wondering if there's also a characteristic class argument that would support this? Or if there's a counterexample that would invalidate the argument?
 A: $\require{AMScd}$
This is a nice qeustion with a nice answer. Your geometric proof looks reasonable though I did not check very hard. Here's an algebraic proof.
The point is that $w_1 \in H^1(\Bbb{RP}^\infty;\Bbb Z/2)$ admits a unique lift to $\tilde w_1 \in H^1(\Bbb{RP}^\infty; \Bbb Z_-)$, where $\Bbb Z_-$ is the nontrivial local system with fiber $\Bbb Z$ (this is a calculation). Now the Poincare duality map $H^1(M;\Bbb Z/2) \to H_{n-1}(M;\Bbb Z/2)$ fits into a commutative square with the Poincare duality map $H^1(M;\Bbb Z_{w_1}) \to H_{n-1}(M;\Bbb Z)$.
\begin{CD}
H^1(M;\Bbb Z_{w_1}) @>PD>> H_{n-1}(M;\Bbb Z) \\
@V\mod 2 VV @V\mod 2VV \\
H^1(M;\Bbb Z/2) @>PD>> H_{n-1}(M;\Bbb Z/2).
\end{CD}
The top-left term is cohomology with local system twisted by $w_1: \pi_1 M \to \pm 1$. Chasing $\tilde w_1$ both ways, we find that $PD(w_1) = PD(\tilde w_1) \pmod 2$, and so $PD(\tilde w_1)$ is the integral homology class / oriented cycle you're looking for.
In general if $R \to S$ is a homomorphism of coefficient systems, then Poincare duality (with suitably twisted coefficients) for $R$ and for $S$ fits into a commutative diagram as above.
