Consider following two infinite groups: group of all permutations of natural numbers (i.e. group of all bijections $f: \mathbb{N} \to \mathbb{N}$) and group of all rotations of a plane. Does group of permutations contain subgroup isomorphic to the group of rotations?
Both groups have cardinality of the continuum, so simple cardinality considerations do not work.
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Edit #2: Here is, to me, an even more surprising result along these lines. Kallmann proved that for any field $F$ of cardinality at most the continuum, $GL_n(F)$ embeds into $S_{\infty}$. In particular, for example, $SO(3)$ has a subgroup of countable index, which is very surprising to me.
Edit: Okay, as suspected the answer to this question is independent of ZF. There's a model of ZF constructed by Shelah in which every set of real numbers has the Baire property. This implies, if I understand correctly, that there are no nonzero homomorphisms from $\mathbb{R}$ to any countable abelian group (since any countable abelian group with the discrete topology is a Polish group, so in this model any homomorphism from $\mathbb{R}$ to such a group is automatically measurable and so automatically continuous). So $\mathbb{R}$, and $SO(2)$, have no subgroups of countable index in this model.
Among other things, in this model $\mathbb{R}$ is a $\mathbb{Q}$-vector space whose $\mathbb{Q}$-linear dual is trivial.
The answer is yes (assuming the axiom of choice; I am quite surprised by this).
More generally, let $A$ be an abelian group and let's see what we can say about the smallest set on which it acts faithfully. If $X$ is a set on which $A$ acts, it breaks up into a disjoint union of orbits $A/A_i$ where the $A_i$ are subgroups of $A$. Because $A$ is abelian, the kernel of $A$ acting on $A/A_i$ is $A_i$, so the kernel of $A$ acting on $X$ is the intersection $\cap_i A_i$.
Specializing to $A = SO(2)$, the question of whether $A$ embeds into $\text{Aut}(\mathbb{N})$ is equivalent to the question of whether $A$ acts faithfully on a countable set, which is in turn equivalent to the question of whether we can find an at-most-countable collection of subgroups $A_i$ of $A$ of at-most-countable index whose intersection is trivial. Now, by the axiom of choice we have an abstract isomorphism
$$SO(2) \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_{i \in I} \mathbb{Q}$$
coming from writing $SO(2) \cong \mathbb{R}/\mathbb{Z}$ and picking a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space containing $\{ 1 \}$. The index set $I$ above is uncountable. By a second application of the axiom of choice, $\bigoplus_{i \in I} \mathbb{Q}$ is abstractly isomorphic to $\mathbb{Q}^{\mathbb{N}}$ (the index set is now countable), so
$$SO(2) \cong \mathbb{Q}/\mathbb{Z} \times \mathbb{Q}^{\mathbb{N}}.$$
Now we can argue as follows. Let $A_i$ be the kernels of the projections to each of the factors $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Q}$ above. Then by construction the $A_i$ are a countable collection of subgroups of countable index, and their intersection is trivial. This means $SO(2)$ acts faithfully on the countable set $\mathbb{Q}/\mathbb{Z} \sqcup (\mathbb{Q} \times \mathbb{N})$ given by the disjoint union of the factors.
(While writing this answer I was repeatedly tempted to conjecture that the intersection of a countable collection of subgroups of countable index has countable index, which is just false, and $\mathbb{Q}^{\mathbb{N}}$ is a counterexample. That's what led to the above construction.)
Of course this argument is deeply inexplicit. Without the axiom of choice I don't know if you can exhibit even a single nonzero homomorphism $SO(2) \to \mathbb{Q}$. All you have is a short exact sequence $0 \to \mathbb{Q}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Q} \to 0$ and it's very unclear what to say about the rightmost term $\mathbb{R}/\mathbb{Q}$ without choice, beyond that it's a $\mathbb{Q}$-vector space.