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Edit: Below I expand my crude original answer "Yes" as requested by the community.

Yes. Let $G$ be the group and $H$ be the closed subgroup. The kernel of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. Hence the kernel is closed. According to this $G/H$ is Hausdorff.

3 added 53 characters in body

Edit: Below I expand my crude original answer "Yes" as requested by the community.

Yes. Let $G$ be the group and $H$ be the closed subgroup. The kernel of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. According to this $G/H$ is Hausdorff.

2 added 436 characters in body

Edit: Below I expand my crude original answer "Yes" as requested by the community.

Yes. The kernel of the quotient map $G \to G/H$ is equal to $\Delta^{-1}(H)$ where $\Delta : G \times G \to G$ is the continuous function $\Delta(x,y)= x- y$. According to this $G/H$ is Hausdorff.

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