A $T_0$ topological vector space is Hausdorff I know a $T_0$ topological group is $T_1$, and if we have a topological vector space, there should be a way of using scalar multiplication to get disjoint neighborhoods. Right?
 A: Yes (although you don't need to use scalar multiplication, since this fact is true for any topological group); hints are below (let $V$ be a topological vector space):
Exercise 1: If for each $0\neq x\in V$, there are disjoint neighborhoods containing $0$ and $x$ in $V$, then prove that $V$ is Hausdorff.
Exercise 2: Let $U$ be a neighborhood of $0$. Prove that $U$ and $x+U$ are disjoint for $x\in V$ if and only if there do not exist $u,u'\in U$ such that $u-u'=x$. 
Exercise 3: If $W\subseteq V$ is a neighborhood of $0$, then prove that there exists a neighborhood $U$ of $0$ such that $U-U:=\{x-y:x,y\in U\}\subseteq V$. (Hint: this is potentially the trickiest past: use the axioms of a topological vector space.)
Exercise 4: Prove that if $V$ is $T_0$, then $V$ is Hausdorff.
Exercise 5: Generalize the ideas above to show that if $G$ is a topological group (e.g., not necessarily abelian), and if $G$ is $T_0$, then $G$ is Hausdorff.
Hope that helps!
A: As suggested in the comments, a topological group which is $T_1$ is already Hausdorff. Here is a proof of this fact.
Remember a space is Hausdorff if and only if the diagonal is closed in the product space. Let $X$ be our space, with operation $\cdot$.
Define $f \colon X \times X \to X$ by $f(x,y)=x\cdot y^{-1}$. Such map is continuous and since the space is $T_1$, $\{\operatorname{id}\}$ is a closed set. The preimage under $f$ of the set $\{\operatorname{id}\}$ is the diagonal, which is closed by continuity, hence $X$ is Hausdorff.
