See page 24 at the top here.
We are supposing the mod $p$ Galois representation $\rho^E_p: G_{\mathbb Q} \to \operatorname{GL}_2(\mathbb F_p)$ is reducible, so that $E$ has a $\mathbb Q$-subgroup $X$ of order $p$. In the case subgroup $X$ is acted on by the trivial character, why are we guaranteed a $\mathbb Q$-point of order $p$? Doesn't every subgroup of order $p$ have a generator, regardless of the action?
As pointed out in the comments, if the representation $G_{\Bbb Q} \to \mathrm{Aut}(E[p])$ is reducible, then there is some subgroup $X \subset E[p] \subset E(\overline{\Bbb Q})$ of order $p$ (1-dimensional over $\Bbb F_p$) which is Galois-stable. This does not mean that $X \subset E(\Bbb Q)$ a priori.
However, if you know that the action of $G_{\Bbb Q}$ on $X$ is trivial, this means that we actually have $X \subset E(\Bbb Q)$, so necessarily $X$ contains a $\Bbb Q$-rational of $E$ of order $p$.