Yes. This is an easy exercise:

For every $r\in\Bbb R$ fix some sequence of rational numbers $r_n$ such that $\lim r_n=r$. Now enumerate $\Bbb Q$ as $\{q_n\mid n\in\Bbb N\}$ and consider $A_r=\{k\mid\exists n:q_k=r_n\}$.

Then given $r\neq r'$, the sequences $r_n$ and $r'_n$ must be disjoint from some point onwards, because $\Bbb R$ is Hausdorff. Therefore $A_r$ and $A_{r'}$ have finite intersection. And so $\{[A_r]\mid r\in\Bbb R\}$ is an antichain of size $2^{\aleph_0}$.

And behold, by choosing $r_n$ to be $q_k$, such that $k$ is the least for which $|q_k-r|<\frac1n$, the entire thing doesn't require a smidgen of choice too!

Yes. This is an easy exercise:

For every $r\in\Bbb R$ fix some sequence of rational numbers $r_n$ such that $\lim r_n=r$. Now enumerate $\Bbb Q$ as $\{q_n\mid n\in\Bbb N\}$ and consider $A_r=\{k\mid\exists n:q_k=r_n\}$.

Then given $r\neq r'$, the sequences $r_n$ and $r'_n$ must be disjoint from some point onwards, because $\Bbb R$ is Hausdorff. Therefore $A_r$ and $A_{r'}$ have finite intersection. And so $\{[A_r]\mid r\in\Bbb R\}$ is an antichain of size $2^{\aleph_0}$.

And behold, by choosing $r_n$ to be $q_k$, such that $k$ is the least for which $0<|q_k-r|<\frac1n$, the entire thing doesn't require a smidgen of choice too!

Yes. This is an easy exercise:

For every $r\in\Bbb R$ fix some sequence of rational numbers $r_n$ such that $\lim r_n=r$. Now enumerate $\Bbb Q$ as $\{q_n\mid n\in\Bbb N\}$ and consider $A_r=\{k\mid\exists n:q_k=r_n\}$.

Then given $r\neq r'$, the sequences $r_n$ and $r'_n$ must be disjoint from some point onwards, because $\Bbb R$ is Hausdorff. Therefore $A_r$ and $A_{r'}$ have finite intersection. And so $\{[A_r]\mid r\in\Bbb R\}$ is an antichain of size $2^{\aleph_0}$.

And behold, the entire thing didn't require a smidgen of choice too!

Yes. This is an easy exercise:

For every $r\in\Bbb R$ fix some sequence of rational numbers $r_n$ such that $\lim r_n=r$. Now enumerate $\Bbb Q$ as $\{q_n\mid n\in\Bbb N\}$ and consider $A_r=\{k\mid\exists n:q_k=r_n\}$.

Then given $r\neq r'$, the sequences $r_n$ and $r'_n$ must be disjoint from some point onwards, because $\Bbb R$ is Hausdorff. Therefore $A_r$ and $A_{r'}$ have finite intersection. And so $\{[A_r]\mid r\in\Bbb R\}$ is an antichain of size $2^{\aleph_0}$.

And behold, by choosing $r_n$ to be $q_k$, such that $k$ is the least for which $|q_k-r|<\frac1n$, the entire thing doesn't require a smidgen of choice too!