Proof:

Since ${\tilde X}_1$ is simply connected the map $p_1$ has a lift $\phi_1:{\tilde X}_1\longrightarrow {\tilde X}_2$ with respect to the covering projection $p_2:{\tilde X}_2\longrightarrow X$, such that $\phi_1({\tilde x}_1) = {\tilde x}_2$. Likewise there exists a lift $\phi_2:{\tilde X}_2\longrightarrow {\tilde X}_1$ of the map $p_2$ with respect to the covering $p_1:{\tilde X}_1\longrightarrow X$, such that $\phi_2({\tilde x}_2) = {\tilde x}_1$. From $p_1\circ \phi_2 = p_2$ and $p_2\circ \phi_1 = p_1$ follows $p_1\circ (\phi_2 \circ \phi_1) = p_1$ and $(\phi_2\circ\phi_1)({\tilde x}_1) = {\tilde x}_1$. Thus, the identity map on ${\tilde X}_1$ and $\phi_2\circ\phi_1:{\tilde X}_1\longrightarrow {\tilde X}_1$ are both lifts of $p_1:{\tilde X}_1\longrightarrow X$ with respect to itself. By uniqueness of lifts we conclude that $\phi_2\circ \phi_1$ is the identity map on ${\tilde X}_1$. Likewise $\phi_1\circ\phi_2$ is the identity map on ${\tilde X}_2$. $\square$