HINT: Your intuition is correct, but it’s probably not the easiest way to attack the problem. Let $X$ be an uncountable set well-ordered by $\le$. For each $x\in X$ let $P(x)=\{y\in X:y\le x\}$, the set of non-strict predecessors of $x$. Let $U=\{x\in X:P(x)\text{ is uncountable}\}$; if $U\ne\varnothing$, let $u=\min U$, and let $X_0=P(u)\setminus\{u\}$; otherwise, let $X_0=X$. In either case $X_0$ is an uncountable set well-ordered by $\le$ such that $P(x)$ is countable for each $x\in X_0$. Let $$L=\{x\in X_0:x\text{ is a limit element}\}\;;$$ you’re done if you can show that $L$ is uncountable.
Suppose that $L$ is countable. Let $A=\bigcup_{x\in L}P(x)$; $A$ is countable, $L\subseteq A$, and if $y<x\in A$, then $y\in A$. That is, $A$ is a countable initial segment of $X_0$ containing all of the limit elements of $X_0$. (These statements need to be justified, of course.) Let $b=\min(X_0\setminus A)$, and get a contradiction by finding a limit element greater than $b$.
(If $L$ has no largest element, you can actually get the contradiction by showing that $b$ is a limit element, but if $L$ has a largest element, $b$ will be its successor, and you really will have to find a limit element strictly greater than $b$.)
