I found Rotman's proof rather fiddly, and my question is why he doesn't do it in the following easier way (probably because I am missing something):
He has already proved (actually, the proof is left to exercises, but these come before his proof of local path connectedness), that if $X$ is a $CW$ complex then
$1),\ $ there is an increasing sequence $\{X^n\}_{n\ge 0}$ of sets whose union is $X$, ($X$ has the weak topology induced by the $X^n), \ X^0$ is discrete and
$2).\ $ for each integer $n$ there is a family of continuous functions $\{f_{\alpha}^{n-1}: S^{n-1}\to X^{n-1}\}_{\alpha\in I}$ such that $X^n=\left(\coprod_{\alpha} D^n\right)\coprod_f X^{n-1}$ where $f=\coprod_{\alpha} f_{\alpha}^{n-1}.$
In particular, $2). $ says that $X^n$ is obtained from a quotient of the disjoint union of $\coprod_{\alpha} D^n$ and $X^{n-1}.$
Now, $X^0$ is (trivially) locally path connected, and it's easy to show that the disjoint union of locally path connected spaces is locally path connected, so if we can show that the quotient of a locally path connect space is locally path connected, then an induction gives what I believe is an alternative, easier proof. Or am I missing something?
But this is not hard:
Suppose $p:X\to Y$ is a quotient map and $X$ is locally path connected. Take an open $V\subseteq Y$ and let $C$ be path component of $V.$ If $x\in p^{-1}(C)$ then $x\in p^{-1}(V)$ also so there is an open path connected $U\subseteq p^{-1}(V)$ containing $x$ (since $X$ is locally path connected).
Now, if $u\in U$ there is a path $\varphi: [0,1]\to U$ such that $\varphi(0)=x$ and $\varphi(1)=u.$ It follows that $p\circ \varphi:[0,1]\to V$ is a path such that $p\circ \varphi(0)=p(x)$ and $p\circ \varphi(1)=p(u).$ But $p(x)\in C$ and this forces $p(u)\in C$ as well, so that in fact $\varphi([0,1])\subseteq p^{-1}(C).$ But then $u\in p^{-1}(C).$
Putting all this together, we see that $U\subseteq p^{-1}(C)$, which means that $p^{-1}(C)$ is open in $X$ and therefore that $C$ is open in $Y$ (since $Y$ has the quotient topology). So $C$ is an open, path connected set contained in $V.$ We conclude that $Y$ is locally path connected.
Your proof looks good to me.
I don't have Rotman's book so I don't know what he does, but I'll say this: I don't think your technique extends to prove the local contractibility of CW complexes. The proof of the latter property is indeed quite a bit more complicated or "fiddly" as you put it, making use of the specific union expressions in properties (2) and (3).