In this answer it is remarked that the real projective plane minus one point is homeomorphic to the Möbius strip without boundary.
A normal Möbius strip is topologically equivalent to a real projective plane with the whole inside of a conic section left out. The conic section has become the single border of the strip. The plane is its single surface. Veblen & Young also describe this equivalence in their books on projective geometry (Vol.2, par. 33).
When we look at a Möbius strip of paper in Euclidean space, we observe two different "mirror versions" of the strip (depending on a right-handed or left-handed twist in the paper).
Question 1:
Is it correct to assume that the projective plane that it models also may be attributed with a certain "mirror version" with respect to its ambient space? As the ambient space, please consider Euclidean space extended with the plane at infinity.
When the conic section shrinks to one point the rest of the plane becomes a Möbius strip without boundary (or: a 1-point boundary).
When I model a Möbius strip by means of paper, then "without boundary" would mean an infinitely large piece of paper but still with the twist.
Question 2:
When question 1 is answered with yes, is it then also allowed to speak of the "mirror version" of a Möbius strip without boundary?
When we say 'without boundary', in this context we mean that the space is not closed (as a subspace of $\mathbb{R}^3$, say), in much the same way that the disc $\{z\mid \|z\|<1\}$ does not include its boundary, but $\{z\mid \|z\|\leq 1\}$ does include its boundary.
Depending on the context, if $X$ is a subset of a metric space then the boundary of $X$ is the closure of $X$ with the interior of $X$ removed. If $X$ is a manifold 'with boundary' then the boundary of $X$ is the set of all points which have a neighbourhood which is homemorphic to the half plane $\{(x,y)\in\mathbb{R}^2\mid y\geq 0\}$ and has no neighbourhood homeomorphic to $\mathbb{R}^2$.