Proving the equivalence of the Criss - Cross theorem statement(ordinary one) to another statement. Here is the Criss - Cross theorem statement:
If $\alpha, \beta : I \rightarrow I \times I$ are paths such that $\alpha(0) = (0,0),\alpha(1) = (1,1),$ and $\beta(0) = (0,1),\beta(1) = (1,0).$ Then there are $a,b \in I $ such that $\alpha(a) = \beta(b).$
And here is the other statement that I want to prove the equivalence to it:
If $A,B \subseteq I \times I$ are path connected subsets such that $(0,0),(1,1) \in A$ and  $(0,1),(1,0) \in B,$ then $A \cap B \neq \emptyset.$
My thoughts:
I know that a space $X$ is path-connected if for every pair of points $a,b \in X,$ there is a path from $a$ to $b$ in $X.$
I know that definition of a path, which is as follows:
A path from $a$ to $b$ in a space $X$ is a continuous function $p: I \rightarrow X$ such that $p(0)= a$ and $p(1) = b.$
Also, I know that a space $X$ is connected if the only separations that $X$ has are trivial separations.
Also, I know the following proposition: 
If $f : X \rightarrow Y$ is continuous and if $X$ is connected then $f(X) \subseteq Y$ is also connected.
I know also, that every path-connected space is also connected. But the converse is not necessarily true.
Even though I know all of the above, I am still unable to prove the equivalence required. How I will pass to the intersection and prove that it is nonempty. Could anyone help me in doing so please?   

$\implies$: since $A$ is path connected and $(0,0),\, (1,1)\in A$, there's a path $\alpha:I\to A\subseteq I\times I$ connecting them. Similarly, there's a $\beta:I\to B\subseteq I\times I$ connecting $(0,1)$ and $(1,0)$.
Then just apply the criss-cross theorem.
$\impliedby$: for given paths $\alpha$ and $\beta$, consider $A:={\rm range}(\alpha)=\alpha(I)$ and $B:={\rm range}(\beta)=\beta(I)$. 
