 Solution

False as a whole: the identity component is normal, but need not be open. Write $G^{\circ }$ for the identity component. The continuous function $(x,y)\mapsto x{y}^{-1}$ maps the connected space $G^{\circ }\times G^{\circ }$ to a connected set containing $e$, hence into $G^{\circ }$. Thus $G^{\circ }$ is a subgroup. Conjugation by any $g$ is a homeomorphism fixing $e$, so it maps $G^{\circ }$ onto itself; therefore it is a normal subgroup. It is also closed, since the closure of a connected set is connected and a connected component is maximal.

For failure of openness, take the additive group $\mathbb{Q}$ with its usual topology. Any two distinct rational numbers can be separated inside $\mathbb{Q}$ by an irrational cut between them, so every connected subset is a singleton. Its identity component is consequently $\{0\}$, which is not open because every neighbourhood of zero contains nonzero rational numbers.