Homotopy Lifting Problem $X, \tilde{X}$ are hausdorff, path connected spaces
$p:\tilde{X} \rightarrow X$ is a local homeomorphism
Under these assumptions we have the following lemma:
If $f:Y \rightarrow X$ is continuous, and $Y$ is connected, f has at most one lift with respect to $p$, that is if $f_1$, $f_2$ are both lifts with respect to $p$, then $f_1 = f_2$ or there is no point in $Y$ where $f_1 = f_2$

I want to prove the following:
Let $a$, $b$ be homotopic paths in $X$ (so that in particular $a$, $b$ have the same starting and endpoints), via the homotopy $F:I \times I \rightarrow X$, and suppose that $f_t = F(.,t)$ each have lifts $\tilde{f_t}$ with respect to $p$ starting at some $\tilde{x_0} \in p^{-1}(a(0))$.
($I$ is $[0,1]$)
Then the paths $\tilde{f_t}$ are homotopic in $\tilde{X}$ and in particular have a common starting and endpoint.
Attempt:
So far I have been able to construct a lifted homotopy, $\tilde{F}:[0,\epsilon] \times I \rightarrow \tilde{X}$ via considering a neighbourhood of $\tilde{x_0}$, $U$ where $p$ is a homeomorphism, then reasoning that $F^{-1}(p(U))$ is open in $I \times I$ and contains ${0} \times I$, so that $[0,\epsilon] \times I \subset F^{-1}(p(U))$ and $F([0,\epsilon] \times I) \subset p(U)$ for some $\epsilon > 0$. Then I defined $\tilde{F} = p \restriction_{U}^{-1} \circ F \restriction_{[0,\epsilon]\times I}$ a lift of $F$ on the desired domain. We also have that $\tilde{F} \restriction_{{0}\times I} = \tilde{x_0}$
So I know that the set of all $\epsilon \in I$ such that a lift as constructed above exists is nonempty, and that each lift corresponding to such an $\epsilon$ is unique, so that any other lift is either a continuation or a restriction of this lift (that is any other lift on some domain $[0,\epsilon_1] \times I$ must agree with lifts on domains that are contained within this domain i.e. $\epsilon_2 < \epsilon_1$) , which follows by our lemma.
I have also been able to show that given such a lift, if $\epsilon < 1$, we can always construct a continuation of the lift, that is we can find a lift $\tilde{F'}:[0,\epsilon + h] \times I \rightarrow \tilde{X}$, where $h$ depends on $\epsilon$,  but don't have control over how much bigger the continuation of the lift is ($h$ is dependent on $\epsilon$). Which implies that if we can show $\sup A \in A$ , then $1 \in A$ and we are done.
I am stuck at this point, as I cannot show that $\sup A \in A$ necessarily.. I would be extremely grateful for any help
EDIT:
I should add I am talking about homotopy of paths all throughout this question, no "free" homotopies where endpoints of each path in the homotopy are different - In this case any lifted homotopy will have to be a homotopy of paths by virtue of the fact that $p$ is a local homeomorphism, and so the preimage of an endpoint is a discrete set.. so this isn't a major concern, what is the primary trouble is how to construct a valid lifted homotopy on all of $I \times I$
 A: If $\tilde{F}(., t) = \tilde{f}_t$ denotes the lifting then it suffices to check that $\tilde{F}$ is continuous into $\tilde{X}$. 
Do it "horizontally" locally instead of "vertically" as you've been trying to do. 
That is, pick a point $(0, t_0) \in I \times I$ and show that there exists an open set $W \subseteq I \times I$ such that $I \times \{ t_0 \} \subseteq W \subseteq I \times I$ such that $\tilde{F}|_{W}$ is continuous. We can now find a "universal" length (which you have been struggling with) doing it horizontally as follows:
Note that for each $\tilde{x} \in \tilde{f}_{t_0}(I \times \{ t_0 \}) = \tilde{F}(I \times \{ t_0 \}) = K$, there exists an open subset $U_{\tilde{x}}$ containing $\tilde{x}$ such that $p(U_{\tilde{x}})$ is open and $p|_{U_{\tilde{x}}}$ is a homeomorphism. Cover $I \times \{ t_0 \} \subseteq \cup_{\tilde{x} \in K} \tilde{f}_{t_0}^{-1}(U_{\tilde{x}})$ and use the Lebesgue covering lemma to find the length $l > 0$ such that if $J$ is an subinterval of $I$ of length at most $l$ then $J \times \{ t_0 \}$ is contained in some $\tilde{f}_{t_0}^{-1}(U_{\tilde{x}})$. We may as well assume that $l = 1/n$ for some positive integer $n$. 
Now the rest is induction with argument similar to yours. First, $\tilde{F}$ is continuous when restricted to some small "vertical" neighborhood of the form $\{ 0 \} \times I_{t_0}$ containing $(0, t_0)$. Pick some $U_{\tilde{x}}$ (as above paragraph) such that $\tilde{F}([0, l] \times \{ t_0 \}) \subseteq U_{\tilde{x}}$. We may cut $I_{t_0}$ if necessary (by using continuity on the "vertical" nbdh $\{ 0 \} \times I_{t_0}$) so that $\tilde{F}(\{ 0 \} \times I_{t_0}) \subseteq U_{\tilde{x}}$. We have $[0, l] \times \{ t_0 \} \in F^{-1}(p(U_{\tilde{x}}))$, so by compactness of $[0,l] \times \{ t_0 \}$ we can fatten $[0,l] \times \{ t_0 \} \subseteq [0,l] \times J_{t_0} \subseteq F^{-1}(p(U_{\tilde{x}}))$ for some neighborhood $J_{t_0}$ of $t_0$. We may cut this further to assume that $J_{t_0} \subseteq I_{t_0}$ so that $\tilde{F}$ is continuous on the first vertical strip $\{0 \} \times J_{t_0}$. Now $\tilde{F} = (p|_{U_{\tilde{x}}})^{-1} \circ F$ on $[0,l] \times J_{t_0}$ by the "unique lifting" lemma that you cited (because both functions start at the same points on the vertical $\{ 0 \} \times J_{t_0}$. The RHS is clearly a continuous expression, so expresses $\tilde{F}$ as a continuous function on a neighborhood of $[0,l/2] \times J_{t_0}'$ where $J_{t_0}'$ is a compact interval containing $t_0$ of length $1/2$ that of $J_{t_0}$. 
Now one simply inducts (go from continuity of $\tilde{F}$ on the strip $\{l \} \times J_{t_0}$ to continuity on $[l, l + l/2] \times T_{t_0}$ for some nondegenerate interval $T_{t_0} \subseteq J_{t_0}$ that could be smaller). Since this stops in finite steps (in at most $2n$ steps to be exact), this "cutting" of the intervals around $t_0$ doesn't pose any problem. The point is that our $l$ doesn't depend on "where" we are on the horizontal strip $I \times \{ t_0 \}$. 
