Is a subgroup of a topological group a topological group?

If $H\le G$ ($H$ is a subgroup of $G$), then the restriction of $*:G\times G\to G$ to $H×H$ is continuous, being the restriction of a continuous map to a subspace (A subtlety which should be appreciated here is that $H×H$ with the product topology is the same as $H×H⊆G×G$ with the subspace topology. Edit: I just realized you only need continuity of $H \times H \to G \times G$, the product of the inclusion $H \to G$ with itself, in order to deduce that $H \times H \to H$ is continuous). The same holds for $()^{-1}:H\to H$.

For the $\overline H$ you need to show that this closure is still a subgroup, i.e. that multiplication of two elements in $\overline H$ does not lead outside of $\overline H$. In other words, show that $*\left(\overline H×\overline H\right)\subseteq\overline H$. The equality $\overline H×\overline H=\overline{H×H}$ may be useful here.

$F \subseteq G$ is a subgroup of G if and only if $a,b \in F$ $\implies$ $ab^{-1}$ is an element of F.

Another way to see this is to prove that the function $f(x,y) = x*y^{-1}$ applied to F is a subset of F.

So we want to show that $f(\overline H×\overline H)$ $\subseteq\overline H$.

We then simply observe f$(\overline H×\overline H)$ $=$ $f(\overline{H \times H})$ $\subseteq \overline {f(H \times H)} \subseteq \overline H$.

Thus $\overline H$ is a subgroup of $G$ and its corresponding multiplication and inversion maps are continuous as restrictions of the corresponding maps of $G$. So $\overline H$ is a topological group.