The answer is yes.
Let $\alpha,\beta:[0,1]\to[0,1]\times \mathbb R$ be two paths; $\alpha(t)=\left(\alpha_1(t),\alpha_2(t)\right)$ and $\beta(t)=\left(\beta_1(t),\beta_2(t)\right)$. Assume that $\alpha_1(0)=\beta_1(0)=0$, $\alpha_1(1)=\beta_1(1)=1$.
Claim. The points $a=\left(0,\alpha_2(0),\beta_2(0)\right)$ and $b=\left(1,\alpha_2(1),\beta_2(1)\right)$ lie in the same connected coponent of the set $\Sigma\subset \mathbb R^3$ formed by points of the following type $ \left(\alpha_1(t),\alpha_2(t),\beta_2(\tau)\right)$ such that $\alpha_1(t)=\beta_1(\tau)$.
Proof. Note that for generic smooth choice of $\alpha$ and $\beta$ the set $\Sigma$ is a smooth 1-dimensional manifold which might be not connected, but it has only two boundary points in $a$ and $b$. Thus, in this case one can connect these points by a curve. The general case can be done by approximation. $\square$
The rest is easy: one can approximate $A$ and $B$ by path-connected sets $A'$ and $B'$ with the same property. Thus one can present $A'$ and $B'$ as a union of curves $\alpha$ and $\beta$ with the above property. Moreover we can assume that the ends of $\alpha$ and $\beta$ are fixed (i.e. $a$ and $b$ are fixed). For each pair $(\alpha,\beta)$, choose the connected component of $a$ in $\Sigma$ and take the union of all of them.