I am thinking about the following statement:
"the union of each collection of nontrivial intervals of $\mathbb{R}$ is the union of a countable subset of that collection"
and I came up with the following explanation:
Let $\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$, where $\mathcal{A}$ is uncountable be such a union of nontrivial intervals of $\mathbb{R}$. Remove from this union every interval that is a subset of the union of other intervals: in this way we get a union of nontrivial intervals $\bigcup_{\alpha\in\mathcal{A'}}I_{\alpha},\ \mathcal{A'}\subset\mathcal{A}$ such that $\bigcup_{\alpha\in\mathcal{A'}}I_{\alpha}=\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$, in each interval we can pick a unique rational number (since each $I_{\alpha}$ is not completely inside any union of other intervals) and from this it follows that $\mathcal{A'}$ is countable, as desired.
Now, I know that another question about the same problem has already been asked, and since the proofs posted there by experienced users are much more complicated I guess there must be something wrong with my reasoning above, but since I don't see what that is and I find it to be an appealing argument I would like to see where it fails and/or/how it could be improved, and, if there are any, other proofs of this interesting statement, thanks.
EDIT 09/19: It might be worthwhile to try by contradiction:
Suppose there existed an uncountable collection of nontrivial intervals $(I_{\alpha})_{\alpha\in\mathcal{A}}$ such that for every countable subset $\mathcal{C}\subset \mathcal{A}$ we have that $\bigcup_{\alpha\in\mathcal{C}}I_{\alpha}\subsetneqq\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$. Then it follows that:
$(I_{\alpha})_{\alpha\in\mathcal{A}}$ cannot contain $(-\infty,\infty)$;
$(I_{\alpha})_{\alpha\in\mathcal{A}}$ cannot contain any countable number of intervals whose union is $(-\infty,\infty)$;
the $(I_{\alpha})_{\alpha\in\mathcal{A}}$ cannot be all disjoint, since in that case it is possible to establish an injection to the rational numbers by picking a distinct rational number from each of them, showing that $\mathcal{A}$ is countable, a contradiction.
(EDIT 09/20) $\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$ cannot be bounded, because from the fact that every non-trivial interval has positive outer measure and $\bigcup_{\alpha\in\mathcal{C}}I_{\alpha}\subsetneqq\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$ for every $\mathcal{C}\subset\mathcal{A}$ countable follows that $|\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}|=\infty$.
So, now the task is to get from these two pieces of information (and perhaps others I haven't yet found out) a contradiction.
Comments are welcome.
NOTE: I have never studied topology, only self-studied calculus/linear algebra/real analysis (and currently measure theory) so I am interested in proofs that don't use too much topological machinery, if that is at all possible.
The main tool to solve the problem is the following result:
This is a particular case of Lindelöf's theorem. The proof is not complicated and I present it here since the OP is not familiar with this result.
Proof: Consider the collection $\mathscr{B}$ of rational intervals ( open bounded intervals with rational endpoints). This is a countable set.
Claim I: Any open interval $I$ in the real line is the union of intervals in $\mathscr{B}$: if $x\in I$ then there is $\delta>0$ such that $(x-2\delta,x+2\delta)\subset I$. Let $q_x\in (x-\delta,x+\delta)\cap\big(\mathbb{Q}\setminus\{x\}\big)$. Choose a rational number $r_x>0$ such that $|x-q_x|<r_x<\delta$. Then $x\in(q_x-r_x,q_x+r_x)\subset (x-2\delta,x+2\delta)\subset I$. Then $I=\bigcup_{x\in I}(q_x-r_x,q_x+r_x)$. Notice that this last union is a countable union since there are $\mathscr{B}$ is countable. $\Box$.
Let $\mathscr{B}'$ be the collection of all rational intervals that are contained in some interval in $\{J_\alpha:\alpha\in\mathscr{A}\}$.
Claim II: $U=\bigcup\{B:B\in\mathscr{B}'\}$. If $x\in U$, then there is $\alpha\in \mathscr{A}$ such that $x\in J_\alpha$. Since $J_\alpha$ is an open interval, Claim I shows there is a rational interval $B_x$ such that $x\in B_x$ and $B_x\subset J_\alpha$. Then $U=\bigcup_{x\in U}B_x$ and so, $U=\bigcup\{B:B\in\mathscr{B}'\}$. $\Box$
For each $B\in\mathscr{B}'$ choose $J_{\alpha_B}\in\{J_\alpha:\alpha\in\mathscr{A}\}$ such that $B\subset J_{\alpha_B}$. The collection of sets $\{J_{\alpha_B}:B\in\mathscr{B}'\}$ is countable since $\mathscr{B}'$ is countable, and $U=\bigcup_{B\in\mathscr{B}'}J_{\alpha_B}$. This concludes the proof of the Theorem. $\Box$.
As in the OP, suppose $\{I_\alpha:\alpha\in\mathcal{A}\}$ is an arbitrary collection of non degenerate intervals, and $$U=\bigcup_{\alpha\in \mathcal{A}} I_\alpha$$ The $I_\alpha$'s may be the form $(a_\alpha,b_\alpha)$,$[a_\alpha, b_\alpha)$, $(a_\alpha,b_\alpha]$, $[a_\alpha,b_\alpha]$ with $-\infty<a_\alpha<b_\alpha<\infty$, or of the form $(a_\alpha,\infty)$, $[a_\alpha,\infty)$, $(-\infty,b_\alpha)$, $(-\infty,b_\alpha]$ with $a_\alpha,b_\alpha\in\mathbb{R}$.
Let $J_\alpha$ be the interior of $I_\alpha$, i.e., the largest open interval contained in $I_\alpha$. For example, if $I_\alpha=[a_\alpha,b_\alpha)$, $J_\alpha=(a_\alpha,b_\alpha)$, etc. Let $V=\bigcup_{\alpha\in\mathcal{A}}J_\alpha$. By the theorem above, there is a countable collection $\{\alpha_n:n\in\mathbb{N}\}\subset\mathcal{A}$ such that $$V=\bigcup_n J_{\alpha_n}$$ If $x\in U\setminus V$ and $x\in I_\alpha$, then $x$ the endpoint of some $I_\alpha$. Let $L$ be the collection of points in $U\setminus V$ that are left endpoints of intervals $I_\alpha$ which are bounded below. Similarly, let $R$ be the collection of all points in $U\setminus V$ that right endpoints of intervals $I_\alpha$ which are bounded above.
If $x\in L$, there is $I_{\alpha_x}$ such that $x\in I_{\alpha_x}$ and $J_{\alpha_x}=(x,b_x)$ where $x<b_x\leq\infty$. There may be several $I_\alpha$'s having $x$ as a left endpoint, but it suffices to choose one for each $x\in L$
Claim: The sets in $\{J_x: x\in L\}$ are pairwise disjoint. Suppose there are $x,x'\in L$, $x<x'$, such that $J_x\cap J_{x'}=(x,b_x)\cap(x',b_{x'})\neq\emptyset$. This implies that $x'\in (x,b_x)=J_x$ and so, $x'\in V$, in contradiction to $x'\in L\subset U\setminus V$. $\Box$
A similar conclusion holds for the points in $R$. For $x\in R$, choose $I_x:=I_{\alpha_x}$ with $x\in I_x$. The sets in $\{J_x: x\in R\}$ are pairwise disjoint.
Claim: The collection $\{J_x: x\in L\}$ is countable. For each $x\in L$, choose a rational number $q_x\in J_x$. Since the $J_x$'s are pairwise disjoint, all the $q_x$'s are distinct. Since the set od rational numbers is countable, it follows that $\{q_x:x\in L\}$ and thus $\{J_x:x\in L\}$ is countable.
A similar result holds for $\{J_x:x\in R\}$.
Let $\{J_n:n\in\mathbb{N}\}$ be an enumeration of the sets that make up $V$; let $\{J'_n: n\in\mathbb{N}\}$ be an enumeration of the sets $\{J_x: x\in L\}$; let $\{J''_n:n\in\mathbb{N}\}$ be an enumeration of the sets $\{J_x:x\in R\}$. Then $\{I_n,I'_n, I''_n:n\in\mathbb{N}\}$ is a countable sub collection of $\{I_\alpha:\alpha\in\mathcal{A}\}$ whose union is also $U$.