Why is this 3-manifold irreducible?

I'm going to extract the essential argument from the general theory of Seifert-fibered spaces. Let $M=\mathbb{R}\mathrm{P}^2\times S^1$, which as a Seifert-fibered space is very special since it is in fact a product $S^1$-bundle over $\mathbb{R}\mathrm{P}^2$.

Let $\alpha\subset\mathbb{R}\mathrm{P}^2$ be a non-separating simple closed curve (an image of a geodesic in $S^2$). Then, $T=\alpha\times S^1\subset M$ is a torus with $M-T$ a solid torus. Let $D=\mathbb{R}\mathrm{P}^2\times\{0\}-T$, which is a disk cross-section of this solid torus, and furthermore let $A=T-\mathbb{R}\mathrm{P}^2\times\{0\}$, which is an annulus. Then, $B=M-(A\cup D)$ is a ball. (Note: I am being cavalier with what I mean by a complement, where to be completely accurate I should be removing tubular neighborhoods.) We can use the sorts of techniques of normal surface theory, as seen in the proof of Kneser's prime decomposition theorem, to put a given surface in a nice position relative to $B$, $A$, and $D$.

Consider an embedded sphere $S\subset M$ that we presume does not bound a ball. By an isotopy, we can assume $S$ is transverse to $A$ and $D$. Suppose $S\cap A$ contains a closed loop that bounds a disk in $A$, and take the innermost such loop and disk. With it, we may do surgery on $S$ to get a pair of spheres. If both spheres bounded balls in $M$, then so too would $S$, so at least one does not; replace $S$ with this sphere. Hence, we may assume $S\cap A$ is a collection of arcs and essential loops in $A$. Similarly, we may assume $S\cap D$ is a collection of arcs.

Consider an innermost arc of $S\cap D$ that bounds a lune in $D$ such that the lune does not contain an antipodal pair of points in $\partial D$ (in the sense that $\partial D$ is a double cover of $\alpha$). We can isotope $S$ along this lune to remove a pair of points of intersection in $S\cap \alpha$. Hence, $S\cap D$ contains only arcs that connect antipodal points of $\partial D$. By considering the $\mathbb{Z}/2\mathbb{Z}$ intersection number, $S\cap D$ contains at most one such arc.

If there were such an arc, $S\cap \alpha$ would be a single point, and so the only arc in $S\cap A$ would be one that connects the two boundary components. This implies there are no essential loops, so $S\cap B$ contains a single component, which must be a disk since $S$ is a sphere. But such a disk in $B$ would imply that $S$ is a torus or Klein bottle, contrary to the fact that it is a sphere. Thus, it must be the case that $S\cap D$ is empty.

It follows that $S\cap A$ is a collection of essential loops. Again, we can assume $S\cap B$ is a collection of disks by performing surgery on $S$ if necessary and keeping one of the two components. Each loop in $S\cap A$, then, corresponds to an $\mathbb{R}\mathrm{P}^2$ component of $S$, so $S\cap A$ is empty.

Hence, $S\subset B$, but Alexander's theorem would imply $S$ bounds a ball, yet $S$ bounds no such thing.