Let $G=\prod_{j=1}^\infty{\mathbb T}$, where ${\mathbb T}=\{z\in{\mathbb C}:|z|=1\}$ is the circle group and $G$ is equipped with the product topology.
Let $H=\prod_{j=1}^\infty\{\pm 1\}$.
Then $H$ is a closed subgroup, but $p:G\to G/H$ is not a fibre bundle.
To see this assume there is a continuous section $s:U\to G$ for some open neighborhood $U$ of $eH$ in $G/H$.
The space $G/H=\prod_{j=1}^\infty{\mathbb T}/\{\pm 1\}$ carries the product topology.
Therefore, $U$ contains an open subset of the form $V=\prod_{j=1}^{N-1}V_j\times\prod_{j=N}^\infty {\mathbb T}/\{\pm 1\}$, where each $V_j$ is an open unit-neighborhood in ${\mathbb T}$.
Let $i:{\mathbb T}/\{\pm 1\}\to \prod_{j=1}^\infty{\mathbb T}/\{\pm 1\}$ be the injection at the $N$-th coordinate, i.e., $i(z)=(1,\dots,1,z,1,\dots)$ with the $z$ in the $N$-th place.
Let $p:\prod_{j=1}^\infty{\mathbb T}\to{\mathbb T}$ the projection onto the $N$-th coordinate.
Then the map $\sigma: {\mathbb T}/\{\pm 1\}\to{\mathbb T}$, given by
$$
\sigma=p\circ s\circ i
$$
is a continuous section to the projection ${\mathbb T}\to{\mathbb T}/\{\pm 1\}$, but such a continuous section does not exist. Contradiction.