I need to prove that the separation axiom $T_0$ is equivalent to $T_1$ in a topological group $G$.
$T_1$ implies $T_0$. So I only need to prove $T_0$ inplies $T_1$.
What I have tried is:
Suppose $G$ is $T_0$, take two elements $x, y \in G$. Then there exists an open set $U$ such that $\left\vert \lbrace x, y \rbrace \cup U \right\vert=1$, without loss of generality $x \in U$. If we use the left translation $\phi_{yx^{-1}} : G \to G$, $\phi _{yx^{-1}} (g)= yx^{-1}g$ then $ \phi _{yx^{-1}}(U)$ is an open set such that $y \in \phi _{yx^{-1}}(U)$. If I prove that $x \notin \phi _{yx^{-1}}(U)$ I finish but I don't know how.
Claim: if $G$ is a topological group and $G$ is $T_0$, then $\{e\}$ ($e$ is the identity element) is closed.
Proof: the last statement is equivalent to: for every $g \neq e$, $G\setminus\{e\}$ is a neighbourhood of $g$.
So pick $g \neq e$. Then $T_0$ implies that either $G\setminus\{e\}$ is a neighbourhood of $g$ (and we are done) or $G\setminus\{g\}$ is a neighbourhood of $e$. So assume the latter. Then $\phi_{g^{-1}}$ (the left translation by $g^{-1}$) is a homeomorphism from $G\setminus\{g\}$ onto $G\setminus\{e\}$, so the set $G\setminus\{e\}$ is a neighbourhood of $\phi_{g^{-1}}(e) = g^{-1}$. Then the map $i$, the inversion map $i(x) = x^{-1}$ is a homeomorphism from $G\setminus\{e\}$ to itself, and shows that $G\setminus\{e\}$ is also a neighbourhood of $i(g^{-1}) = g$, as required.
In this case, if $\{e\}$ is closed, all singletons are closed by homogeneity (or directly using translations), so $G$ is $T_1$.
In your proof, I don't yet see how to show $x \notin \phi_{yx^{-1}}[U]$, which would indeed be enough. E.g. in the reals with the usual topology, take $x=0, y=1$, and take $U = (-2, \frac{1}{2})$. Translating $0$ to $1$ does leave $0$ also in the translated open set as well. So your proof cannot work without picking $U$ more carefully.