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 $\Sigma$.