If a subgroup has finite-intersection with a compact neighbourhood of $1$, then it is discrete?

Question

At page 79 of Basic Number Theory by André Weil, there is an argument showing that a subgroup of a topological group is discrete,

because there is a compact neighbourhood of $1$ with finite intersection with that subgroup $\Gamma$.

So I am wondering how one proves the following:

If a sub-group of a topological Hausdorff group has finite intersection with a compact neighbourhood of $1$, then it is discrete.

Since one has earlier spotted the statement that a discrete subset of a compact set is finite, I think, in this book, by discrete one understands closed discrete.
Now, we know that this subset $\Gamma$ has a discrete intersection with a compact neighbourhood. But how could this fact help us showing the discreteness of the whole subset $\Gamma$? This is where I am stuck.
Any hint is well-appreciated.

Answer 1

Let me put the comments of Alex Youcis together to see if I understand rightly.
Firstly, we need only to find one neighbourhood of $1$ that does not contain any other elements of $\Gamma$, since we are working with a topological group.
Then we use the Hausdorff-ness of the group to find neighbourhoods $U_i$ of $1$ such that $x_i\not\in U_i \forall x_i\in\Gamma\cap V, x_i\not=1,$ where $V$ is the neighbourhood in question. And the desired neighbourhood is just $\bigcap U_i\cap V,$ a finite (number of ) intersection(s) by assumption.
Indications of errors are mostly welcomed.

Answer 2

Alex's comments, as turned into an answer by the OP, are indeed correct.

Let me just point out that the Hausdorff condition really is the key here. For instance, let $G$ be any group admitting a nontrivial finite subgroup $H$. Endow $G$ with the indiscrete topology (in which the only open subsets are $\varnothing$ and $G$): this makes $G$ into a non-Hausdorff topological group. But the desired conclusion fails: $G$ is a compact neighborhood of the identity which has finite intersection with the subgroup $H$, and $H$ is not discrete (rather it is indiscrete).