We assume $A$ is the set of all countable subsets of the set of real numbers. We know $A$ is a partially ordered set $(A, \subseteq)$.
Suppose $$A_1 \subseteq A_2 \subseteq \ldots \subseteq A_n \subseteq A_{n+1} \subseteq \ldots$$ is a chain in $A$. We can prove $B=\bigcup_{n \in \Bbb{N}} A_n$ is a countable set.
For each natural number $m$, we have $A_m \subseteq B$. So $B$ is an upper bound for $A$. This shows each chain in $A$ has an upper bound according to Zorn's lemma. $A$ has a maximal element $X$, and we know $X$ is a countable set. Now we prove $X = \Bbb{R}$.
If $X \neq \Bbb{R}$, then there is an $x \in \Bbb{R}$ such that $x \notin X$. Let $Y=X \cup \{x\}$. It's obvious that $Y$ is a countable subset of the real numbers and $X \subsetneq Y$. This contradicts $X$ being a maximal element.
Thus, $X = \Bbb{R}$ and $\Bbb{R}$ is a countable set.
What is wrong with this argument?