= Solution
Choose $\widehat x_i\in p_i^{-1}(x_0)$. By the <classification of connected covering spaces> and the <degree of a connected covering>, each cover corresponds to an index-two subgroup
$$
H_i=(p_i)_*\pi_1(\widehat X_i,\widehat x_i)
\le\pi_1(X,x_0)\cong\mathbb Z^2.
$$
An index-two subgroup is the kernel of a nonzero homomorphism $\mathbb Z^2\to\mathbb Z/2\mathbb Z$. There are three:
$$
\{(m,n):m\equiv0\pmod2\},\qquad
\{(m,n):n\equiv0\pmod2\},\qquad
\{(m,n):m+n\equiv0\pmod2\}.
$$
The group $GL_2(\mathbb F_2)$ acts transitively on the three nonzero linear functionals on $\mathbb F_2^2$, and its elementary matrices lift to the <integral general linear group>. Hence there is $A\in GL_2(\mathbb Z)$ such that
$$
AH_1=H_2.
$$
Take $\psi=f_A$. Since $\psi_*(H_1)=H_2$, the <lifting criterion for a covering space> gives a based lift
$$
\phi:\widehat X_1\to\widehat X_2
$$
such that
$$
\boxed{p_2\circ\phi=\psi\circ p_1}.
$$
Applying the same argument to $A^{-1}$ gives a lift in the reverse direction. The composites are based lifts of the appropriate identity maps; uniqueness of based lifts makes them identities. Thus $\phi$ and $\psi$ are homeomorphisms, and the required square commutes. This proves the <equivalence of connected double covers of the torus>.
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