You don't. $A\subseteq B$ and $B\supseteq A$ means the same thing. The same thing goes for proper subset and supersets. The first chain means:
$$I_n\subset...\subset I_2\subseteq I_1$$
while the second means
$$I_n\subseteq...\subseteq I_2\subseteq I_1$$
The first implies the second so you can't have the first to be true without having the second.
If you meant it the second the other way around you would now that they're contradicting. If $A\subset B$ then $A\subseteq B$ is definitely false.
If you meant using non-proper subsets all the way it becomes somewhat interresting.
So for a concrete case where $I_n = [1/n, 2]$ you can easily see that $I_j\supset I_k$ if $j>k$ because the implication that $x\in I_k$ implies that $x\in I_j$ in combination that you can find $x\in I_j$ that's not in $I_k$ (the later makes the superset proper). You only have to consider the definitions for the internvals to see this (especially that $1/k\le x$ implies that $1/j\le x$ if $j>k$ and for being proper you can consider $1/j\le x=1/j$ but $1/k> x=1/j$).
You could also apply the nested interval theorem too, which implies that $I_j\supset I_k$ if $j>k$ immediately (bot not the converse). The proof above is basically similar to the proof of the nested interval theorem.
Note that you could have the both ways, but that means that all $I_j$ are equal (that's why you need to make them non-proper).