I think you mean $\supseteq$ and not $\subseteq$, since otherwise $\displaystyle \bigcap_{n \in \mathbb{N}} I_n = I_1$.
Assuming you mean $\supseteq$, if you had $\mathbb{R}$ instead of $\mathbb{N}$, then the hypothesis would presumably be that $I_r \supseteq I_s$ for all $r<s$. In this case, you'd have
$$\bigcap_{x \in \mathbb{R}} I_x ~=~ \bigcap_{n \in \mathbb{N}} I_n$$
since $I_x \supseteq I_{\mathrm{max}\{1, \lceil x \rceil\}}$ for each $x \in \mathbb{R}$.
So this property for $\mathbb{N}$-indexed intervals implies the same property for $\mathbb{R}$-indexed intervals.
Let $(a, b)$ be an open interval, and $(c, d)$ an open interval properly contained in $(a, b).$ Then $a \leqslant c < d \leqslant b,$ and $a < c$ or $d < b.$ The set $(a, b) \setminus (c, d) = (a, c] \cup [d, b)$ is not open, because it contains $c$ or $d$ or both, but it does not contain a neighbourhood of either. Therefore $(a, b)$ is not the disjoint union of $(c, d)$ with the union of any non-empty collection of open intervals.
An application
(Proposition 4 is the promised application, while Proposition 5 is a
by-product of the argument.)
Proposition 1. An open interval is not the disjoint union
of an open interval and a non-empty open set.
Proof. See above. $\ \square$
Proposition 2. The union of a non-empty collection of open
intervals with a non-empty intersection is an open interval.
Proof. Let $\mathscr{I}$ be a non-empty collection of open
intervals containing a given point $c \in \mathbb{R},$ and let
$J = \bigcup\mathscr{I}.$ In
$\overline{\mathbb{R}} = \mathbb{R} \cup \{+\infty, -\infty\},$ let
$a = \inf J$ and $b = \sup J.$ Then $a \notin J$ and $b \notin J.$
If $a < x < b,$ then $c \leqslant x < b$ or $a < x \leqslant c,$ and
in either case $x \in I \subseteq J$ for some $I \in \mathscr{I}.$
Therefore $J = (a, b).$ $\ \square$
Proposition 3. If $x \in U \subseteq \mathbb{R},$ and $U$
is open, then $U = J \cup W,$ where $x \in J,$ $J$ is an open
interval, $W$ is an open set, and $J \cap W = \varnothing.$
Proof. Let $J$ be the union of all open intervals $I$ such
that $x \in I \subseteq U.$ By Proposition 2, $J$ is an interval
$(a, b).$ Clearly, $a \notin U$ and $b \notin U,$ therefore
$$
U = (a, b) \cup (U \cap (-\infty, a)) \cup (U \cap (b, +\infty)),
$$
so we can take
$W = (U \cap (-\infty, a)) \cup (U \cap (b, +\infty)).$ $\ \square$
Proposition 4. An open interval is not the disjoint union
of two non-empty open sets.
Proof. Let $I$ be an open interval, and suppose that
$I = U \cup V,$ where $U$ and $V$ are disjoint non-empty open sets.
Take any $x \in U.$ By Proposition 3, $U = J \cup W,$ where
$x \in J,$ $J$ is an open interval, $W$ is an open set, and
$J \cap W = \varnothing.$ Therefore
$$
I = (J \cup W) \cup V = J \cup (W \cup V), \text{ and }
J \cap (W \cup V) = (J \cap W) \cup (J \cap V) = \varnothing.
$$
This contradicts Proposition 1; so the hypothesis that
$I = U \cup V$ is false. $\ \square$
Proposition 5. Every open subset of $\mathbb{R}$ is the
union of a countable collection of pairwise disjoint open intervals.
Proof. Let $U$ be an open subset of $\mathbb{R},$ and let
$\mathscr{J}$ be the collection of all maximal open subintervals of
$U.$ By Proposition 3, $U = \bigcup\mathscr{J},$ and any two members
of $\mathscr{J}$ with a non-empty intersection are equal. Because
each member of $\mathscr{J}$ contains a rational number,
$\mathscr{J}$ is countable. $\ \square$
Best Answer
Open intervals are still fine by this definition. In less compact language, the definition states that an interval $I$ is composed of all the numbers such that if you pick two numbers $x$ and $y$ in the interval, then every $z$ between them (including endpoints) is also in the interval. So, if you have an open interval, meaning the endpoints are not included, then you can't take either of the endpoints as your $x$ and $y$.
As for the set $[3,2]$, remember that $$[3,2] = \{x\in \mathbb{R} : 3 \leq x \leq 2\}$$ by definition of the notation $[a,b]$. However, in this case, as you've pointed out, $[3,2] = \emptyset$. However, we can still think of it as the empty interval, because it is vacuously true that if you pick any two numbers $x,y$ in the empty set then every number in between those two numbers are also in the empty set (because no two such number exist in the first place).