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
Let us begin with the definition of an interval on real line: a subset $I$ of the real line is an interval if and only if for any two points $a<b$ in $I$ and any $a<c<b$, one has $c\in I$.
Note that the union $T$ of a sequence of open intervals $I_n$ is automatically open. So what remains is to check that $T$ is still an interval. Let $a,b$ be any two point in $T$ and suppose that $a<b$. The two points $a$ and $b$ must lie in some given intervals, say $a\in I_1$ and $b\in I_2$. Let $c$ be any point in $\mathbb{R}$ such that $a<c<b$. But as $I_1\cap I_2$ is not empty, so one can take $d\in I_1\cap I_2$ which satisfies $a<d<b$ (there are only finitely possibile situations for two open sets to intersect and you can compute it out explicitly by considering the sup and inf of the intervals). If $c=d$, then it is done as $c\in I_1\cap I_2 \subset T$; otherwise either $a<c<d$ or $d<c<b$ will imply that $c\in I_1\cup I_2\subset T$ as $I_1$ and $I_2$ are intervals. The proof is complete.
The assumption of mutually overlapping is quite strong. In fact it is much stronger than what is needed for the union to remain an interval. For more information, you can check about connectedness or even just path-connectedness in topology.
If you are not familiar with topology or you are not sure about why an (arbitrary) union of open sets is still open, you can check it by considering the sup and inf of this union, once you succeed in proving that it is an interval.