 Solution

False. Give the two-element group $C_2$ the indiscrete topology and take $G=\mathbb{R}\times C_2$ with the product topology. The indiscrete factor is a topological group, since every map into an indiscrete space is continuous, and the product group operations are continuous. Points $(0,e)$ and $(0,a)$, $a\ne e$, have identical neighbourhoods, so $G$ is not a Hausdorff space. However, its continuous projection onto $\mathbb{R}$ is surjective. If $G$ were compact, its image $\mathbb{R}$ would be compact, which it is not.