The proof in Dummit & Foote is the familiar induction proof. $\require{amssymb}$
$G$ is an abelian group. Let $N = \langle x \rangle$ for some $x \in G$. Since $G$ is abelian, $N \trianglelefteq G$. We use induction to conclude that $G \, / \, N$ has an element $\bar{y} = yN$ of order $p$, where $p$ is prime, since $|G \, / \, N| < |G|$ and $p$ divides $|G|$ but $p$ does not divide $|N|$.
My question is, if $|yN| = p$ and all cosets have the same order, why can't we conclude $|yN| = |N| = |x| = p$?
Or, more accurately, how can we say $p$ does not divide |$N$|, then show that $p = |yN|$. Isn't that a contradiction?
You are using two different concepts of order. Yes, $yN$ has order $p$ as an element of $G/N$. That does not mean that $\#yN=p$. What it means is that $y^pN=N$; in other words, $y^p\in N$.