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 $\Sigma$.