Prove Locally Euclidean Topological Group is Hausdorff?

Question

I am self-studying differential and general topology, and am trying to show the following:

Let $G$ be a locally Euclidean topological group. Prove $G$ is Hausdorff.

Using this, I understand why the space would be Hausdorff if the points $p,q\in G$ have a neighborhood $U$ containing both points. However, I am struggling to prove that if there does not exist a Euclidean neighborhood containing both points, the space will still be Hausdorff. I know it has to do with the fact that the elements form a group, and I think it has to do with the group property of continuous inversion and product. I have been trying for awhile to write this proof and have been unable to get anywhere. I have tried using local maps and homomorphisms of $G$ onto $(\mathbb{R}^n,+)$, but I have not been able to get anything to work. I imagine a proof by contradiction could be effective (assume $\nexists\hspace{1mm}U,V\in(G,\tau)\hspace{1mm}s.t.\hspace{1mm}U\cap V=\varnothing$), and I imagine you would then suppose a point in $U\cap V$ and use group characteristics to show a contradiction. Is this a correct line of reasoning? Could I have a pointer in the right direction for what group properties I'd need to consider, or if a different approach would be better?

UPDATE:

Using the method given in the answer, here is the proof:

$1)\hspace{1mm}\forall x\in\mathbb{R}^n, \mathbb{R}^n\backslash\{x\}=(-\infty,x)\cup(x,\infty)\therefore \{x\}$ is closed for all singletons, and $\mathbb{R}^n$ is $T_1$.

$2)$ Given $x,y\in G,\exists U,V\in(G,\tau)\hspace{1mm}s.t.\hspace{1mm}x\in U,y\in V,x\notin V,y\notin U$. By definition of the group product ($U\times V\rightarrow W$ and $(x,y)\rightarrowtail xy^{-1}$) being continuous, $xy^{-1}$ must have an open neighborhood $W$. Suppose $j\in U\cap V$. Then, $(j,j)=jj^{-1}=e$ must be in the neighborhood of $W$. Other way around, if $e\in W$, then there must be a shared point in $U$ and $V$, because $e=jj^{-1}=(j,j)$. Because $G$ is $T_1$, the neighborhood $W$ can be restricted to not include $e$, thus restricting the preimage of the product, which means $\exists U,V\in(G,\tau)$ such that $x\in U, y\in V$, and $U\cap V=\varnothing$.

Answer 1

1. Prove that every locally Euclidean space is T1, i.e. every singleton set is closed.

2. Prove that a topological group is T1 iff it is T2 (i.e. Hausdorff).

The proof for 1) is straight forward. Just reduce the general case to the special case of euclidean space. The proof of 2) takes two distinct elements $x,y$ and then considers the element $x * y^{-1}$.