This is exercise 6-4 on page 181 in John Lee's Topological Manifolds book which asks me to prove the above, that is, if $M$ is a boundaryless surface which contains a subset $B$ which is homeomorphic to the Möbius band, then $M$ is of the form $M'\#\mathbb RP^2$ where $M'$ is a boundaryless surface.
I tried to attack it in this way:
Try to find a polygonal presentation for $M$ which has a (possibly non-adjacent) twisted pair $aa$. I know how to make it adjacent by means of cutting and pasting (this process is outlined in part I of the proof of the classification theorem). Then we have $M$ presented as $Vaa$ for some word $V$. So set $M'$ to be the surface presented by word $V$ and we are done because concatenation of words gives the presentation of the connected sum and $aa$ presents $\mathbb RP^2$.
The non-trivial part is to find that crucial twisted pair. For this I thought, first triangulate $M$. Then cut along $B$, a Möbius band sitting in $M$. Do we get a polygonal presentation of $M\setminus B$?
Later I looked at the hint which asks us to show that there is $B_0\subsetneq B$ which is homeomorphic to the Möbius band. That's clear to me. But why consider this subset at all?
I know this counts as two questions, but the aim is to just solve this problem.
Thanks in advance!
The answer was almost given by amomin, let us just formulate it precisely.
Take $B_0\subset B$, homeomorphic to the Möbius Band, that you have seen.
Glueing a disk to $M\setminus B_0$ yields a boundaryless surface $M'$. Then, $B_0$ is homeomorphic to $\mathbb{R}P^2$ minus a disc (see $\Bbb RP^2$ as the union of a Möbius band and a disc ).
Removing the disk to $M'$, then glueing $\mathbb{R}P^2$ minus a disc corresponds exactly to do $M' \# \mathbb{R}P^2$, and the result is $M$.