https://en.wikipedia.org/wiki/Homotopy_lifting_property

video

$p: E \to B$ is a Covering, $p(e_0) = b_0$. Let $F: I \times I \to B$ ($I=[0,1]$) be continuous, with $F(0\times 0) = b_0$. Then there is a lifting $\tilde{F}: I \times I \to E$ s.t. $\tilde{F} (0\times 0) = e_0$ (it's also unique).

If $F$ is a Path homotopy (constant on the sides $\{0\}\times I$ and $\{1\} \times I$), then also $\tilde{F}$ is a path homotopy.

Proof. Uses Lebesgue number lemma to take an open covering of $B$, which gives also an open covering of square (which is compact and metric), given by the preimages $V$ of the Covering. Then we can divide $I \times I$ into squares whose images are contained in the open sets of the covering of $B$. (see proof of Path lifting lemma for details).