Lemma 1
Let $f\colon X \rightarrow X'$ and $g\colon Y \rightarrow Y'$ be open maps of topological spaces.
Then $f\times g \colon X\times Y \rightarrow X'\times Y'$ is open.
Proof:
Clear.
Lemma 2
Let $X, Y$ be topological spaces.
Let $R$(resp. $S$) be an equivalence relation on $X$(resp. $Y$).
Suppose the canonical map $X \rightarrow X/R$(resp. $Y \rightarrow Y/S$) is open.
Let $R\otimes S$ be the equivalence relation on $X\times Y$ such that $(x, y) \equiv (x', y')$ (mod $R\otimes S)$ if and only $x \equiv x'$ (mod $R)$ and $y \equiv y'$ (mod $S)$.
Then the canonical map $(X\times Y)/(R\otimes S) \rightarrow (X/R)\times (Y/S)$ is an isomorphism.
Proof:
Let $f\colon X\times Y \rightarrow (X/R)\times (Y/S)$ be the canonical map.
By Lemma 1, $f$ is open.
Hence $f$ induces an isomorphism $(X\times Y)/(R\otimes S) \rightarrow (X/R)\times (Y/S)$.
QED
Lemma 3
Le $X$ be a topological space.
Let $\Delta$ be the diagonal of $X\times X$, i.e, $\Delta = \{(x, x) \in X \times X\colon x \in X\}$.
Then $X$ is Hausdorff if and only if $\Delta$ is closed.
Proof:
Left to the readers.
Lemma 4
Le $X$ be a topological space.
Let $R$ be an equivalence relation on $X$.
Suppose the canonical map $X \rightarrow X/R$ is open.
Let $C$ be the graph of $R$, i.e. $C = \{(x, y) \in X\times X\colon x \equiv y$ (mod $R)\}$.
Suppose $C$ is closed.
Then $X/R$ is Hausdorff.
Proof:
Let $f\colon X\times X \rightarrow (X/R)\times (X/R)$ be the canonical map.
By Lemma 2, $f$ induces an isomorphism $(X\times X)/(R \otimes R) \rightarrow (X/R)\times (X/R)$.
Let $\Delta$ be the diagonal of $(X/R)\times (X/R)$.
Since $f^{-1}(\Delta) = C$, $\Delta$ is closed.
Hence $X/R$ is Hausdorff by Lemma 3.
QED
Proposition
Let $X$ be a topological group.
Let $G$ be a closed subgroup of $X$.
Let $X/G$ be the space of left cosets of $X$ by $G$.
Then $X/G$ is Hausdorff.
Proof:
Let $R$ be the equivalence relation on $X$ defined by $x \equiv y$ (mod $R)$ if and if $x^{-1}y \in G$.
Then $X/G = X/R$.
Let $C$ be the graph of $R$.
Let $f\colon X\times X \rightarrow X$ be the map definded by $f(x, y) = x^{-1}y$.
Then $f$ is continuous and $f^{-1}(G) = C$.
Since $G$ is closed, $C$ is closed.
Clearly the canonical map $X \rightarrow X/G$ is open.
Hence $X/G$ is Hausdorff by Lemma 4.
QED