Can I use Poincaré duality to prove that the first homology group of a non-orientable manifold $M$ is not zero? I only need to prove that $H_1(M;\mathbb Z_2)\ne0$, and by the universal coefficient theorem I can deduce $H_1(M)\ne0$.
So how to show that $H_1(M;\mathbb Z_2)\ne0$?
You can do this by Poincare duality. The details are in Hatcher (Cor 3.28). The idea is that for a non-orientable closed n-manifold $H_n(M;\mathbb{Z})=0$ but $H_n(M;\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$. Then universal coefficients show that the torsion subgroup of $H_{n-1}(M;\mathbb{Z})$ equals $\mathbb{Z}/2\mathbb{Z}$. Hence $H_{n-1}(M;\mathbb{Z}/2\mathbb{Z})\not=0$. Now Poincare duality (with $\mathbb{Z}/2\mathbb{Z}$ coefficients) tells us that $H_1(M;\mathbb{Z}/2\mathbb{Z})$ is non zero.
If you like vector bundles, you can also use that since $M$ is not orientable, there exists a non-trivial line bundle on $M$. This is de determinant of the tangent bundle. Real line bundles are in correspondence with cohomology classes with $\mathbb{Z}/2\mathbb{Z}$ coefficients. Hence $H_1(M;\mathbb{Z}/2\mathbb{Z})\not=0$.