We are asked to prove that if $G$ is a topological group, and $H$ is a subspace of $G$ that is also a subgroup, then $H$ and $\bar{H}$ (the closure of $H$) are topological groups. We have to do a few extra proofs to demonstrate that this is true:
First, we must show that is some function $f$ is continuous, then resticting its domain will still yield a continuous function. We can do this be realizing that for some continuous function $f:X \ \rightarrow \ Y$, if we restrict the domain to $S \ \subset \ X$, then the new function $f_s \ = \ f \ \circ \ i$, where $i$ is the continuous inclusion map from $X$ to $S$. Thus, $f_s$ is continuous. This means that if the functions $f(x) \ = \ x^{-1}$ and $g(x \ \times \ y) \ = \ x \ \otimes \ y$ are continuous in $G$, then they will be continuous for the subspaces $H$ and $\bar{H}$.
Next, we must show that if some topological space has the propoert of the $T_1$ axiom, then so do its subspaces. The $T_1$ axiom says that any finite point set is closed. This implies that for each pair of points in our space, $x$ and $y$, each point has a neighbourhood that doesn't contain the other one. We can quickly prove this. Let us say that each finite point set is closed. This means $\{x\}$ is closed $\forall x \ \in \ X$, where $X$ is our topological space. Then $X \ - \ \{x\}$ is an open negihbourhood for every other point in $X$ that doesn't contain $x$. Conversely, if every point $x \ \in \ X \ - \ \{y\}$ has a neighbourhood that doesn't contain $y$, we can take the union of all of these neigbourhoods to find that $X \ - \ \{y\}$ is open, making $\{y\}$ closed. We can then take finite unions of these single point sets to get other closed sets, proving that finite point sets are closed.
Getting back to the original point, if we want to prove that for each $x \ \in \ A$, where $A$ is our subgroup has a neighbourhood not containing $y$, we simply have to recall that we have already imposed the $T_1$ axiom on $X$, thus for the points $x$ and $y$, there is some open $U$ in $X$ such that $x \ \in \ U$ and $y \ \notin \ U$. Since $x, \ y \ \in \ A$, it follows that $x \ \in \ U \ \cap \ A$ and $y \ \notin \ A \ \cap \ U$. We know that $A \ \cap \ U$ is open in $A$, thus we have proven the $T_1$ axiom holds.
The only thing that is left to prove now is that $\bar{H}$ is a subgroup, and to do that, we will require some fancy footwork.
We want to prove that $\overline{H \ \times \ H} \ = \ \bar{H} \ \times \ \bar{H}$ Let $x \ \times \ y \ \in \ \overline{H \ \times \ H}$. It follows that every open neighbourhood around $x \ \times \ y$ intersects $H \ \times \ H$. Among those open neighbourhoods are basis elements of the form $U \ \times \ V$, for every pair $U$ and $V$ open in $H$. Thus, we have $U \ \cap \ H \ \neq \ \emptyset$ for every neighbourhood $U$ of $x$, and $V \ \cap \ H \ \neq \ \emptyset$ for every neighbourhood $V$ of $y$. This means that $x \ \in \ \bar{H}$ and $y \ \in \ \bar{H}$, so $x \ \times \ y \ \in \ \bar{H} \ \times \ \bar{H}$. Conversely, let $x \ \times \ y \ \in \ \bar{H} \ \times \ \bar{H}$. This means that every neighbourhood of $x$ and every negihbourhood of $y$ intersects $H$. This means that $x \ \times \ y \ \in \ U \ \times \ V$. Since $U \ \times V$ forms a basis for open set in $H \ \times \ H$, if we have $x \ \times \ y \ \in \ U \ \times \ V$ that intersects $H \ \times \ H$, then this will be true for any neighbourhood $W$ around $x \ \times \ y$, as we can always chooe $U \ \times \ V$ with $x \ \times \ y \ \in \ U \ \times \ V \ \subset \ W$. This means that each $W$ will intersect $H \ \times \ H$, and $x \ \times \ y \ \in \ \overline{H \ \times \ H}$. Since we have demonstrated inclusion both ways, it follows that $\bar{H} \ \times \ \bar{H} \ = \ \overline{H \ \times \ H}$.
Also, recall from earlier in the textbook that if some function $f$ is continuous, then $f(\bar{A}) \ \subset \ \overline{f(A)}$. Now, we can jump into the actual proof. Since the functions $f(x) \ = \ x^{-1}$ and $g(x, \ \times \ y) \ = \ x \ \otimes \ y$ are continuous in $X$, we will have:
$$f(\bar{H}) \ = \ \bar{H}^{-1} \ \subset \ \overline{f(H)} \ = \ \overline{H^{-1}} \ \subset \ \bar{H}$$
Since the inverse of each element of a group uniquely exists and is contained within the group, we have $H^{-1} \ \subset \ H$. We have thus shown that $\bar{H}^{-1} \ \subset \ \bar{H}$, which means that the inverse of each element of $\bar{H}$ is contained within $\bar{H}$. Since $H \ \subset \ \bar{H}$, we know the identity is in $\bar{H}$, and associativity follows from the nature of the $\otimes$ operation. Finally, we have:
$$g(\bar{H} \ \times \ \bar{H}) \ \subset \ \overline{g(H \ \times \ H)} \ \subset \ \bar{H}$$
This means that the $\otimes$ operation between any two elements of $\bar{H}$ will be contained within $\bar{H}$. We have thus satisfied all the axioms necessary for $\bar{H}$ to be a group! $\blacksquare$