The lifting theorem 

This is The Lifting Theorem for covering spaces proven in Bredon's book Topology and Geometry. 
There is just one thing I don't get how it works.
Namely, in the last paragraph the author states that one has to show that g is continuous.  
What confuses me is, which definition exactly is he using to prove continuity?
At first I thought he takes any nbh. $U$ of $g(w)$ and wants to find a nbh. $V$ such that the whole $g(V)$ is contained in that nbh. $U$, but then it seems like he proves (only) that for $V$ given in the proof $g(V)$ is contained in some nbh. of $g(w)$ whose image under the covering map $p$ is a trivialising nbh. for point $f(w)$ in $X$. 
What am I missing and why is this enough?
 A: We want to prove that $g: W \to X$ is continuous. It suffices that check that $g$ is continuous locally, i.e. it suffices to check that for every $w \in W$, there exists an open neighbourhood $U'$ of $g(w)$ and an open neighbourhood $V$ of $w$ such that $g(V) \subset U'$, and such that $g|_{V} : V \to U'$ is continuous.
To show this is the case, recall that $p : X \to Y$ is a covering map. This means that there exists an open neighbourhood $U$ of $f(w) = p(g(w))$ such that $p^{-1}(U)$ is a disjoint union of open sets, all of which are homeomorphic to $U$ via the projection $p$. Now, $g(w)$ is contained in one of these open sets. We define $U'$ to be precisely this open set.
Next, we take $V$ to be any path-connected open neighbourhood of $w$ contained within $f^{-1}(U)$. [Such an open neighbourhood exists because $W$ is locally path-connected.] By the construction of the map $g$, we have $g(V) \subset U'$.
Finally the map $g|_V : V \to U'$ is precisely the map $f|_V : V \to U$ composed with the homeomorphism $p^{-1} : U \to U'$. Again, this more-or-less follows from the definition of $g$: the discussion about "concatenation of paths" in the final paragraph in the book aims  to explain this point.
But $f|_V$ and $p^{-1}$ are both continuous, so their composition $g|_V$ is continuous.
