Continuity of the lift ${\tilde f}$:

Let $y \in Y$ be arbitrary, and let $f(y) = x$ and ${\tilde f}(y) = {\tilde x}$. Choose an evenly covered neighborhood $U$ of $x$ and ${\tilde U}$ be the sheet containing ${\tilde x}$ lying above $U$. By continuity of $f$ we obtain a neighborhood $V$ of $y$ in $Y$ such that $f(V) \subset U$ and hence ${\tilde f}(V) \subset p^{-1}(U)$ (since $p\circ {\tilde f} = f$).

Now if we assume that ${\tilde f}$ maps the neighborhood $V$ into ${\tilde U}$, then the following would be valid:

$${\tilde f} = \Big(p\Big\vert _{\tilde U}\Big)^{-1} \circ f, \eqno(18.2)$$

which would prove the continuity of ${\tilde f}$. To prove that ${\tilde f}(V) \subset {\tilde U}$, we shall assume that the neighborhoods $U$, $V$ and ${\tilde U}$ are path connected and invoke the construction of ${\tilde f}$. Choose a path $\gamma$ in $Y$ joining $y_0$ and $y$ and for each $z \in V$ pick a path $\eta$ joining $y$ and $z$ and then we get the path $\gamma *\eta$ joining $y_0$ and $z$. Lift $f \circ \gamma$ and $f \circ \eta$ to paths in $\tilde X$ starting at ${\tilde x}_0$ and $\widetilde{f \circ \gamma}(1)$ respectively. Since $f \circ \eta$ lies in $U$, its lift must lie entirely in $\tilde U$ and hence

$${\tilde f}(z) = \widetilde{f \circ (\gamma*\eta)}(1) = \widetilde{f \circ \eta}(1) \in \tilde U.$$