= Solution
\b[False as a whole: the <identity component> is normal, but need not be open.] Write $G^\circ$ for the <identity component>. The <continuous function> $(x,y)\mapsto xy^{-1}$ 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.
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