The idea is sound, but the implementation could be better. Here’s a fairly careful write-up missing only a few details.
Let $U$ be a non-empty open subset of $\Bbb R$, and let $\mathscr{I}$ be the family of all open intervals contained in $U$. For $I,J\in\mathscr{I}$ write $I\sim J$ iff there are $I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$; clearly $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-equivalence class of $I$, and let $U_I=\bigcup[I]$; clearly $U_I$ is open.
Suppose that $U_I\cap U_J\ne\varnothing$ for some $I,J\in\mathscr{I}$; then there are $I'\in[I]$ and $J'\in[J]$ such that $I'\cap J'\ne\varnothing$. Clearly $I\sim I'\sim J'\sim J$, so $I\sim J$, and $[I]=[J]$. Thus, $\{U_I:I\in\mathscr{I}\}$ is a partition of $U$ into open subsets.
Suppose that $I,J\in\mathscr{I}$ and $I\cap U\ne\varnothing$; then $I\cup J\in\mathscr{I}$. (Why?) An easy induction on $n$ then shows that if $I_0,\dots,I_n\in\mathscr{I}$, and $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$, then $\bigcup_{k=0}^nI_k\in\mathscr{I}$. Fix $I\in\mathscr{I}$, and suppose that $x,y\in U_I$ with $x<y$. Then there are $I_x,I_y\in[I]$ such that $x\in I_x$ and $y\in I_y$. Now $I_x\sim I\sim I_y$, so there are $I_x=I_0,\dots,I_n=I_y\in[I]$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Let $J=\bigcup_{k=0}^nI_k$; then $x,y\in J\in[I]$, so $[x,y]\subseteq J\subseteq U_I$. This shows that $U_I$ is order-convex: if $x,y\in U_I$ and $x<z<y$, then $z\in U_I$.
The order-convex subsets of $\Bbb R$ are precisely the intervals, so the open, order-convex subsets of $\Bbb R$ are the open intervals (including the open rays), so $\{U_I:I\in\mathscr{I}\}$ is a partition of $U$ into open intervals, as desired.
You can also prove it by defining an equivalence relation directly on $U$: for $x,y\in U$, $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. This makes some of the argument a little easier and some a little harder; overall it’s probably pretty much a wash.
One possible approach is to write down the most general interpretation of the definition, and see what happens. A countable union of disjoint open sets is a set of the form
$$ \bigcup_{n=1}^{\infty} U_n $$
where $U_m\cap U_n = \emptyset$ whenever $m\ne n$ and each $U_n$ is open.
Note that the emptyset itself is open and that the definition does not require that the sets in the union be nonempty. So, for example, we can write
$$ (0,1) = \bigcup_{n=1}^{\infty} U_n, $$
where $U_1 = (0,1)$ and $U_n = \emptyset$ for all $n > 1$. You should check that this really is a disjoint union of open sets, and that it really gives you $(0,1)$, but this check should be pretty straight-forward.
Alternatively, if we understand the word "countable" to mean "any natural number or countably infinite" then the union
$$ \bigcup_{n=1}^{1} (0,1) $$
gets the job done.
As per the comments, a part of the question that I did not address is the following:
Do there exist two (or more) disjoint open sets $U_1$ and $U_2$ such that $(0,1) = U_1 \cup U_2$?
The answer is "No." The interval $(0,1)$ is a connected set (see this question for an argument that justifies this statement; we lose no generality replacing the closed unit interval with the open unit interval). If there were two such open sets, then they would form a disconnection of the interval $(0,1)$, which is a contradiction. Note that the non-existence of a nontrivial disjoint cover of $(0,1)$ by open sets does not violate the original theorem in any way, via the reasoning given in the first part of this answer.
Best Answer
Let In$[a,b]$ denote the closed interval from $a$ to $b.$ That is In$[a,b]=[a,b]\cup [b,a].$ This removes the need to distinguish the cases $a<b,a=b, a>b.$
For open $ E\subset \mathbb R $ and $ a,b\in E $ let $ a\sim b $ iff In$[a,b]\subset E.$ Obviously $ a\sim b$ iff $b\sim a, $ and also $ a\sim a, $ for any $ a,b\in E.$
Exercise. For any real $a,b,c$ we have $$In[a,b]\cup In[b,c]=In [\min (a,b,c),\max (a,b,c)]=[\min (a,b,c),\max (a,b,c)].$$ Corollary : $\sim$ is transitive.
So $\sim$ is an equivalence relation on $E.$
Exercise.The set of equivalence classes of any equivalence relation on any set comprise a partition of that set.
Note that in your title we implicitly include open half-lines, the whole real line, and the empty set among the open intervals. The members of the partition $E_{/\sim}$ are convex open sets (open because if $x\in E$ then $(x-d,x+d)\subset E$ for some $d>0,$ so $(x-d,x+d)\subset [x]_{\sim}$ ),... but not necessarily bounded open intervals. For example if $E=(1,\infty)$ then $E_{/\sim}=\{E\}.$
If $x,y\in \mathbb R$ with $x<y$ and $(x,y)\in E_{\sim}$ then neither $x$ nor $y$ belongs to $E$.
Another way of describing $E_{/\sim}$ is : For $a\in E, $ let $a^*$ be the set of convex open real set(s) $T$ such that $a\in T\subset E.$ Let $a^{**}=\cup a^*.$ Then $a^{**} \in a^*,$ so $a^{**}$ is the $\subset$-maximum member of $a^*.$ And $a^{**}=[a]_{\sim},$ the $\sim$-equivalence class containing $a.$