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?
In order to use Zorn's lemma, you need to show that every chain has an upper bound. You have only shown that countable chains of the form $E_1\subseteq E_2\subseteq E_3\subseteq \cdots$ have an upper bound.
I cannot think of a constructive example of a chain in $A$ with no upper bound, but there is a nonconstructive one. Let $f:\omega_1\to \mathbb R$ be an injective function, where $\omega_1$ is the first uncountable ordinal number, equal to the set of countable ordinal numbers. Then the collection $\{f(A):A\in \omega_1\}$ is a chain, whose upper bound has cardinality $|f(\omega_1)|=|\omega_1|>\aleph_0$. Essentially, $f$ is a well-ordered, uncountable sequence of real numbers where every initial segment is countable.
Since $|\omega_1|\le |\Bbb R|$, such an injection $f$ exists, but I do not think we can construct such an $f$.