 Solution

Each left coset $gH$ is nonempty, since it contains $g=ge$, and every $g\in G$ lies in its own left coset. Suppose $aH$ and $bH$ intersect, say $a{h}_1=b{h}_2$ with $h_1,h_2\in H$. ThenSince $H$ is a subgroup, $h_2{h}_1^{-1}\in H$, and $aH\subseteq bH$. The reversed relation $b=a{h}_1{h}_2^{-1}$ gives $bH\subseteq aH$. Thus any two left cosets are either equal or disjoint, and together they cover $G$. Therefore $G/H$ is a partition of a set $G$. No finiteness or normal subgroup assumption is needed.