If $X$ is a contractible space, then for every neighborhood $W$ of $*$, there is a neighborhood $U$ of $*$ such that $U \subseteq W$ and $U$ is contractible in $W$.
My book's proof:
Let $\text{id} \simeq_F *: X \to X$. Then $F(*,t) = * \in W$ and so by continuity of $F$, there exist neighborhoods $U_t$ of $*$ and $V_t$ of $t$ such that $F(U_t \times V_t) \subseteq W.$ Clearly $\{V_t\}_{t \in I}$ covers $I.$ By compactness of $I$, there is a finite sub cover $V_{t_1}, \ldots, V_{t_n}$, and we set $U = \bigcap U_{t_i}.$ It follows that $F|U \times I: U \times I \to W$ is the desired nullhomotopy of the inclusion map.
Question
Firstly, why is it clear that $\{V_t\}_{t \in I}$ covers $I$? Does cover mean span? Secondly what does the "$|$" symbol mean when they write $F|U \times I: U \times I \to W$?
"Cover" means "cover". Just to show two examples, it might be that $V_{0.5}=(0.4,0.7)$ and $V_{0.9}=(0.75,1]$ (note that the first one contains $0.5$ and the second one contains $0.9$; that's what $V_t$ bring a neighbourhood of $t$ signifies). The main point is that every point $a\in I$ is contained in some $V$ (specifically $V_a$), which means that the infinite union of all $V_t$ is the whole unit interval.
The vertical bar means "restricted to". If you have any function $f:A\to B$ and $A'\subseteq A$ is any subset, then $f|_A$ is the function that takes as input points $a\in A'$ and send it to $f(a)\in B$.