Can a countable group have uncountably many subgroups? If $G$ is a countable group,can it have an uncountable number of distinct subgroups?
Let $V$ be a vector space of dimension $\aleph_0$ over a countable field $F$ (so $V$ is countable) and let $B$ be a basis for $V$ over $F$. Then every subset of $B$ spans a different subspace of $V$, so $V$ has $2^{\aleph_0}$ different subspaces, and its additive group has $2^{\aleph_0}$ different subgroups.
For each $(b_n) \in \prod_{n\in \mathbb{N}}(\mathbb{Z}/2)$ consider the map from $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$ to $\mathbb{Z}/2$ 
$$(a_n) \mapsto \sum_n a_n \cdot b_n$$

take the kernel of this map; we get $2^{\aleph_0}$ subgroups of index $2$ of $\oplus _{n\in \mathbb{N}}(\mathbb{Z}/2)$.

@user1729: thanks! for  correcting a previous statement about the number of subgroups of finite index of a countable group -- it may well be uncountable! 

The example of @seirios: is a finitely generated group having uncountably many subgroups. 
However, a finitely generated group has finitely many subgroups of a given index $n$. Indeed, one can reduce to normal subgroups ( the normal core has index $N \le n!$) and then notice that there are finitely many morphisms from the groups to the finitely many groups of order $N$. Thus a finitely generated group has countably many subgroups of finite index.