At the bottom of page $64$ of Hatcher it says:
The preceding paragraph implies that $p: \widetilde{X} \rightarrow X$ is continuous.
We've already shown that the map $p|_{U_{[\gamma]}}:U_{[\gamma]} \rightarrow U$ is a homeomorphism, where we've defined $$U_{[\gamma]} = \{[\gamma\cdot \eta]\mid \eta \text{ is a path in }U \text{ with }\eta(0)=\gamma(1)\}$$
Why is $p: \widetilde{X} \rightarrow X$ continuous?
You are defining $p$ gluing together many homeomorphisms along a basis of open sets; this in general leads to a continuous map.
This is the general result.
Gluing Lemma. Let $Y$ be a topological space and let $\mathfrak{U}:=\{U_i\}_{i\in I}$ be an open cover of a topological space $X$. Suppose $\{f_i:U_i\longrightarrow Y\}_{i\in I}$ is a family of continuous functions with the following property: for every $i,j\in I$ such that $U_i\cap U_j\neq \varnothing$ we have $$f_i|_{U_i\cap U_j}\equiv f_j |_{U_i\cap U_j}$$ Then there exists an unique continuous map $f:X\longrightarrow Y$ such that $f|_{U_i}\equiv f_i$ for each $i\in I$.