Let $\mathcal A$ be a family of infinite countable sets which is linearly ordered by inclusion and such that $\bigcup\mathcal A$ is uncountable. I have to prove that $|\bigcup\mathcal A|=\aleph_1.$
I think I can prove it when $\mathcal A$ is well ordered. Let $\kappa=\mathrm{type}(\mathcal A).$ Then we can index $\mathcal A$ with ordinal numbers $\iota<\kappa:$ $\mathcal A=\{A_\iota\}_{\iota<\kappa}.$ If $\kappa>\aleph_1,$ then there is $A_{\aleph_1}.$ But this is the union of all $A_\iota$ over $\iota<\aleph_1$. Since $A_\iota$ are pairwise distinct, $A_{\aleph_1}$ must be an uncountable set, a contradiction with $A_{\aleph_1}\in\mathcal A.$ Therefore the ordinal type of $\mathcal A$ is at most $\aleph_1.$ But if it were less than that, then $\bigcup\mathcal A$ wouldn't be uncountable.
But I can't see how I can do it for a linearly ordered $\mathcal A.$ Is there some kind of classification for linear orders like the one that ordinal numbers provide for well-orders? (Please, even if it's irrelevant, do answer this question, or tell me that I should ask it separately.)
Your idea of looking at well-ordered sets is the key. It is true that $\mathcal A$ needs not be well-ordered, but its cofinality is well-defined anyway.
Here, the cofinality of $\mathcal A$ is simply the least ordinal $\alpha$ such that there is a strictly increasing sequence of length $\alpha$ of elements of $\mathcal A$ that is cofinal in $\mathcal A$, that is, a sequence $\langle A_\beta\mid \beta<\alpha\rangle$ such that each $A_\beta\in\mathcal A$, if $\beta<\gamma<\alpha$, then $A_\beta\subset A_\gamma$, and for any $A\in\mathcal A$, we have $A\subset A_\beta$ for some $\beta<\alpha$ (and $\alpha$ is least possible).
Now note that this $\alpha$ is a cardinal, at most $\aleph_1$ (it could be $1$), and that $\bigcup A=\bigcup_{\beta<\alpha}A_\beta$, and the problem reduces to the well-ordered case.
In fact, we do not even need to talk of cofinalities here, all that matters is that there is such a sequence $\langle A_\beta\mid\beta<\alpha\rangle$, regardless of whether $\alpha$ is least or not. Now of course $\alpha$ is not necessarily a cardinal, but still $\alpha\le\aleph_1$, which is all that is needed.