Every point in g has an open neighborhood that is evenly covered. These neighborhoods cover the path, which is compact, so take a finite subcover.
These open neighborhoods map back, through g, to open sets that cover I. Replace each open set with base open sets, i.e. open intervals in I. Our finite cover has become infinite again, but I is compact, so a finite subcover will do. Now I is covered by finitely many open intervals, and the image of each, under g, is evenly covered by p.
Let a closed interval C1 cover most of the first open interval O1, crossing into the second interval O2. (You can prove that the "lowest" open interval O2, not contained in O1, has to intersect O1, else the boundary of O1 is not covered.) So C1 is contained in O1 and pokes its nose into O2. Let C2 pick up where C1 left off, covering most of O2, and reaching into O3. Continue this process, untill there are a finite number of closed intervals covering I, and each maps to a subpath of g that is evenly covered by p.
Look at the first subpath g(C1). This is contained in an evenly covered open set U1. There may be many copies of U1 in the preimage; select the one that contains x. Then apply the homeomorphism in reverse to map the path in the base space back up to a path in the total space. Call this path f. On the first closed subinterval C1, p(f) = g. We just need to take the next step, and the next.
By induction, assume g has a unique lift over the first n-1 subintervals. Now g(Cn) lives in an evenly covered set Un, and there is only one preimage of Un in S~ that contains the end point of f(Cn-1), so select that copy and apply the homeomorphism to extend f across Cn.
Since g(Cn) and f(Cn) are homeomorphic, and g is continuous, f is continuous on this closed interval. Is there any continuity trouble at the point of connection? For the entire path up through Cn, p(f) = g. This holds throughout our open set Un. Thus f and g are homeomorphic for all of Cn, and the tail end of Cn-1, as dictated by the open set Un. Since g is continuous across this juncture, so is f.
Run through all the closed sets in I, and the unique lift is complete. Our function f in S~ is a continuous path, and p(f) = g throughout.