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 $\alpha_1(0)=\beta_1(0)=0$, $\alpha_1(1)=\beta_1(1)=1$ and $0<\alpha_1(t),\beta_1(t)<1$ for $0< t< 1$.

The principle part is to show that the points $a=\left(0,\alpha_2(0),\beta_2(0)\right)$ can be conneced to $b=\left(1,\alpha_2(1),\beta_2(1)\right)$ in 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 $\alpha_1(t)=\beta_1(\tau)$. The rest can be done by passing to smaller sets and applying the fact that any connected set in the plane can be approximated by path-connected sets

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.

In general case can be done by approximation; i.e. we get that $a$ and $b$ are in the same connected component of $\Sigma$.

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.

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 $\alpha_1(0)=\beta_1(0)=0$, $\alpha_1(1)=\beta_1(1)=1$ and $0<\alpha(t),\beta(t)<1$ for $0< t< 1$.

The principle part is to show that the points $(0,0,0)$ can be conneced to $(1,1,1)$ in 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 $\alpha_1(t)=\beta_1(\tau)$. The rest can be done by passing to smaller sets and applying the fact that any connected set in the plane can be approximated by path-connected sets

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 $(0,0,0)$ and $(1,1,1)$. Thus, in this case one can connect these points by a curve.

In general case can be done by approximation; i.e. we get that $(0,0,0)$ and $(1,1,1)$ are in the same connected component of $\Sigma$.

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 $\alpha_1(0)=\beta_1(0)=0$, $\alpha_1(1)=\beta_1(1)=1$ and $0<\alpha_1(t),\beta_1(t)<1$ for $0< t< 1$.

The principle part is to show that the points $a=\left(0,\alpha_2(0),\beta_2(0)\right)$ can be conneced to $b=\left(1,\alpha_2(1),\beta_2(1)\right)$ in 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 $\alpha_1(t)=\beta_1(\tau)$. The rest can be done by passing to smaller sets and applying the fact that any connected set in the plane can be approximated by path-connected sets

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.

In general case can be done by approximation; i.e. we get that $a$ and $b$ are in the same connected component of $\Sigma$.