You'll be able to understand this more clearly if you keep track of base points. Once that's done, they key is to apply, over and over, the uniqueness properties of lifting theory, which you use to tell you that this map is equal to that map.
Let's fix base points $x \in X$ and a lifted base point $\tilde x \in X$. Let $\tilde x' = \tilde p(\tilde x)$, also a lift of $x$.
The map $\tilde p : \tilde X \to \tilde X$ is the unique map (by covering theory) which is a lift of $p : \tilde X \to X$ such that $\tilde p(x)=x'$.
There also exists (by covering theory) a unique map $\tilde p_1 : \tilde X \to \tilde X$ which the lift of the map $p : \tilde X \to \tilde X$ such that $\tilde p_1(x')=x$.
The composition $\tilde p_1 \tilde p : \tilde X \to \tilde X$ is a lift of the map $p$ such that $\tilde p_1 \tilde p(x)=x$. The identity map $\mathbb{1}_{\tilde X}$ satisfies the same property: it is a lift of the map $p$ such that $\mathbb{1}_{\tilde X}(x)=x$. The map characterized by this property is unique (again by covering theory) and so $\tilde p_1 \tilde p = \mathbb{1}_{\tilde X}$.
A similar argument shows that $\tilde p \tilde p_1 = \mathbb{1}_{\tilde X}$.
Thus, $\tilde p$ and $\tilde p_1$ are inverses of each other, hence each is a homeomorphism.
Now, as to the issue of showing that $\tilde p = \mathbb{1}_{\tilde X}$, you've not included enough information for that purpose. The lifting lemma guarantees uniqueness of a lift for a given point and a given image of that point. And, as you have reported the proof, that information is not given. If you could structure the proof with specifically named points and specifically named images of those points, and if you can do this so that $\tilde p$ fixes some point of $\tilde X$, then uniqueness of lifting would guarantee that $\tilde p = \mathbb{1}_{\tilde X}$.
I thought for a while now and can not detect any flaws in the following, so here we go:
Pick $y \in Y$. We may, as you said, pick some open neighbourhood of $y$ say $N_y$ and a natural number $n$ s.t. $H(N_y \times [ \frac{k-1}{n},\frac{k}{n}])$ lies inside an evenly covered neighbourhood $U_k$. Say $(V_{k,i})_{i\in I}$ are the disjoint open sets that map homeomorphically to $U_k$ via $\pi$.
Now here comes my reasoning why I think we can skip the connectedness of $N_y \times \{ \frac{k}{n} \}$: $\tilde{f}(y)$ lies in one of the $V_{1,i}$, and after replacing $N_y$ by $\tilde{f}^{-1}(N_y)$ we may assume that so does all of $\tilde{f}(N_y)$. However in this case one can define a continous lift $H'_{1}$ of $H|_{N_y \times [0, \frac{1}{n}]}$ simply by composing with $\pi^{-1}$. Since by construction the whole image of this lift lies in one of the $V_{1,i}$ we can repeat this process (with $\tilde{f}$ replaced by $H'_{1}(-,\frac{1}{n})$) to construct a continous lift $H'$ of $H|_{N_y \times [0, 1]}$. Moreover as you wrote this continous lift has to coincide with $\tilde{H}|_{N_y \times [0, 1]}$. After all for every $z \in N_y$, $H'(z, -)$ and $\tilde{H}(z,-)$ both provide continous lifts of $H(z, -)$ with starting point $\tilde{f}(z)$ and path-lifting is always unique.
Best Answer
To apply the homotopy lifting property, you need to assume that $f: X\to B$ has a lifting. It doesn't assert that any homotopy can be lifted without assumptions. Your first bullet point is not correct.
However all 3 facts stated are correct (modulo some extra conditions on the topological spaces).
For a simple example highlighting why your argument doesn't hold, consider $f:\mathbb S^1\to\mathbb S^1$ the identity map, $\pi:\mathbb R\to \mathbb S^1$ the universal covering map. Then $f$ cannot be lifted, to $\tilde f:\mathbb S^1\to \mathbb R$, hence any homotopy between $f$ and another map cannot be lifted as well.