= Solution
Let $\widetilde\alpha$ and $\widetilde\beta$ be lifts to the compactified <hyperbolic plane> $\overline{\mathbb H^2}$. Suppose they have two common points. Choose two consecutive common points along one lift. If both lie in $\mathbb H^2$, the intervening subarcs contain an innermost embedded disc. The covering projection is injective on its interior; otherwise a nontrivial deck translate would produce a still smaller such disc. Its projection is a bigon between $\alpha$ and $\beta$, contrary to the hypothesis.
The same innermost-disc argument works when one corner is on the circle at infinity, after deleting a sufficiently small <horoball> about that corner. The only possible obstruction would identify the ideal corner by both a hyperbolic deck transformation associated with an essential return and a parabolic deck transformation stabilizing the relevant puncture. A hyperbolic and a parabolic element of the surface group cannot have that fixed point in common, so the truncated disc again projects to a bigon. If both common points are ideal, truncate at both ends and apply the same argument. Therefore any pair of lifts intersects in at most one point of $\overline{\mathbb H^2}$.
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