Is there a proper subgroup $H$ of $(\mathbb{R},+)$ such that $[\mathbb{R}:H]$ is finite?

We claim that there is no subgroup of finite index in $\mathbb R$. Here is the proof:

Proof:

Suppose $\mathbb R$ has a proper subgroup of finite index. That is there exists a proper subgroup $H$ such that, $[\mathbb R:H]=n$.

Since, $\mathbb R$ is an abelian group, $H$ is a normal subgroup. So, the group of cosets of $H$ in $\mathbb R$ has the property that $(rH)^n=H$ for each $r\in \mathbb R,$ which implies that $nr \in H $ for each $r \in \mathbb R.$

Note that, this property is a consequence of the Lagrange's Theorem: The order of an element in a finite group divides the order of the group.

From this, we claim that, $\mathbb R \subset H$, which will contradict that $H$ is a proper Subgroup of $\mathbb R$.

For each $r \in \mathbb R$, $\dfrac{r}{n} \in \mathbb R$. Now, we have, $n \cdot \dfrac{r}{n}=r \in H$. This proves the claim and hence the contradiction.

Aside:

This in fact proves something more: As ymar points to in his comments: Any divisible group cannot have proper subgroups of finite index. Now what are these groups: Intuitively, those groups in which you can divide! Formally, these are groups in which each element is an $n^{th}$ mutiple of some element for each $n \in \mathbb N$.

This observation is implicit in the other answer.

Let $A$ be a domain which is not a field, and let $M$ be a finitely generated divisible $A$-module. Then $M=0$.

Assume by contradiction $M\neq0$, and let $N$ be a maximal proper submodule of $M$. Such exists because $M$ is nonzero and finitely generated. Then $S:=M/N$ is isomorphic to $A/\mathfrak m$, where $\mathfrak m$ is a maximal ideal of $A$. As $A$ is not a field, $\mathfrak m$ contains a nonzero element $a$, and we have:

$\bullet\ $ $0\neq S=aS$, because $S$ is divisible and $a$ is nonzero,

$\bullet\ $ $aS=0$, because $a$ is in $\mathfrak m$.