(Hatcher- Algebraic topology) p.30
To prove (c) we will first construct a lift $\overline{F}:N\rightarrow \mathbb{R}$ for $N$ some neighborhood in $Y$ of a given point $y_0 \in Y$. Since $F$ is continuous,every point $(y_0,t)\in Y×I$ has a product neighborhood $N_t\times(a_t,b_t)$ such that $F:N_t\times(a_t,b_t) \subset U_\alpha$ for some $\alpha$. By compactness of $\{y_0\}×I$, finitely many such products $N_t\times (a_t,b_t)$ cover $\{y_0\}×I$. This implies that we can choose a single neighborhood $N$ of $y_0$ and a partition $0=t_0 <t_1 <···<t_m =1$ of I so that for each $ i, F(N\times [t_i,t_{i+1}])$ is contained in some $\alpha$.
I don't get the last assertion.
I will illustrate how I followed this argument.
Let $\{U_i\}$ be the collection of evenly covered sets by $p$.
Hence, $\{F^{-1}(U_i)\}$ is a cover of $Y$. Since $\{y_0\}\times I$ is compact, by Tube lemma, there exists an open neighborhood $V_y$ of $\{y_0\}$ such that $V_y$ is covered by finite subset of $\{F^{-1}(U_i)\}$.
Define $\gamma(t)=F(y_0,t)$
Apply Lebesgue'a number lemma, then there is a partition $\{p_0,...,p_m\}$ of $[0,1]$ such that $ \gamma ([p_i,p_{i+1}])$ is contatined in some $\alpha$.
However, as the bolded line says, how do I construct the neighborhood $N$?