What needed to be checked for a topological subgroup?

The path-connected component of a point $x$ in some space $X$ consists of all points $y\in X$ such that there exists a $xy$-path in $X$. All those $y$ define a subset and thus with the subspace topology a subspace.

A topological group is a topological space $G$ together with a chosen identity element $e_G \in G$ and two continuous maps $\cdot:G \times G \rightarrow G$ and $(-)^{-1}:G \cong G$ satisfying the usual group axioms (associativity of multiplication, unitality of $e_G$ and having inverses given by $(-)^{-1}$).

You are supposed to show that the connected component subspace of the element $e_G$ of a topological group $G$ again forms a topological group. $e_G$ is contained in its own connected component, because there is a constant $e_Ge_G$-path. It is not so obvious that given $x,y$ in the component $x\cdot y$ is in the component. There we need to recall that a $xy$-path is given by a continuous map $[0,1] \rightarrow X$ sending $0$ to $x$ and $1$ to $y$. Since both $x$ and $y$ are in the component of $e_G$ there are $xe_G$- and $ye_G$-paths $\alpha,\beta$. Now the multiplication is a continuous map $G\times G \rightarrow G$ and we can use the paths $\alpha,\beta$ to define a continuous map $$\begin{array}{ccccc} [0,1] & \overset{(\alpha,\beta)}{\longrightarrow} & G \times G &\overset{\cdot}{\longrightarrow} & G\\ t & \mapsto & (\alpha(t),\beta(t)) & \mapsto & \alpha(t)\cdot \beta(t)\\ 0 & \mapsto & (x,y) & \mapsto & x\cdot y\\ 1 & \mapsto & (e_G,e_G) & \mapsto & e_G \cdot e_G = e_G \end{array}$$ This defines a $(x\cdot y)e_G$-path proving that $x\cdot y$ is contained in the component of $e_G$.

I leave the proof of $x^{-1}$ being in the component up to you. The idea is precisely the same: use an $xe_G$-path and continuity of $(-)^{-1}$ to define an $x^{-1}e_G$-path.

If $a,b\in G_0$ and $\gamma\colon [0,1]\to G$ is a path from $id$ to $a$, then $[0,1]\to G$, $t\mapsto \gamma(t)b$ is a path from $b$ to $ab$. Append this to a path from $id$ to $b$ to find a path from $id$ to $ab$. Also note that $[0,1]\to G$, $t\mapsto a^{-1}\gamma(1-t)$ is a path from $id $ to $a^{-1}$. Conclude that $G_0$ is a subgroup.

Use a similar simple argument for normality (or perhaps exhibit a homomorphism to a suitable group that has precisely $G_0$ as kernel - this could kill subgroup and normality in a single strike).