Mapping class group vs outer automorphism group of the fundamental group for nonorientable surfaces

Part of the theorem is true in much more generality. Let $M$ be an Eilenberg-MacLane space of type $K(\pi, 1)$. That $Out(\pi_1(M))$ is isomorphic to $\pi_0 HomotopyEquivalences(M)$ is a standard fact -- presumably Farb and Margalit's argument factors through this argument? I haven't read their book so I don't know the answer but I suspect it's yes. It appears in Hatcher's Algebraic Topology notes, anyhow.

To finish the proof for surfaces you have to argue that for non-orientable surfaces $\pi_0 Diff(M) \to \pi_0 HomotopyEquivalences(M)$ is an isomorphism of groups. This breaks into two questions -- (1) is every homotopy-equivalence homotopic to a diffeomorphism, and (2) can non-trivial diffeomorphisms be homotopic to the identity?

Looking at the Margalit-Farb paper it looks like they answer (2) by lifting the diffeomorphism to a map of the universal cover, in the case that's a hyperbolic plane (the non-hyperbolic cases are considered special cases), and then extend to the disc compactification. This is a fairly standard type of argument.

For (1) they use what they call the Dehn-Nielsen-Baer theorem, but they also give alternative proofs, later in Theorem 8.9. One nice way to think of this is to do it inductively -- homotope the image (under the homotopy-equivalence) of a simple closed curve to an embedding. Then you can cut the domain and range along those curves, and reduce the problem to one for a lower-genus surface, where the homotopy-equivalence is a diffeomorphism on the boundary already. Eventually you get to a homotopy-equivalence of discs which restricts to a diffeo on the boundary (you need to cut along arcs to do this) and then at that point you can use the straight-line homotopy, since discs are convex.