Here is Prob. 3 in the Supplementary Exercises, Chap. 2, in the book Topology by James R. Munkres, 2nd edition:
[Let $G$ be a topological group.] Let $H$ be a subspace of $G$. Show that if $H$ is also a subgroup of $G$, then both $H$ and $\overline{H}$ are topological groups.
Here is the link to an earlier post of mine on topological groups here on Math SE.
My Attempt:
As $G$ is a topological group, so $G$ is a topological space that satisfies the $T_1$ axiom and thus each singleton subset of $G$ is closed in $G$. [See Sec. 17 in Munkres.] Therefore each singleton subset of $H$ can be written as the intersection set of that set with $H$ and is thus closed in the subspace $H$. So the subspace $H$ is also a topological space that satisfies the $T_1$ axiom.
As $G$ is a topological group, so the map $f \colon G \times G \to G$, defined by $$ f(x \times y) \colon= xy^{-1} \ \qquad \mbox{ for all } x \times y \in G \times G, \tag{A} $$ is continuous. [Visit the above link.]
As $H \subset G$, so $H \times H \subset G \times G$, by Prob. 2 (j), Sec. 1, in Munkres.
Let $g$ denote the map obtained from $f$ by restricting the domain of $f$ to the subset $H \times H$ of $G \times G$. Then, by Theorem 18.2 (d) in Munkres, $g$ is also continuous.
As $H$ is a subgroup of $G$, so, for any elements $x, y \in H$, the element $xy^{-1} \in H$ also. That is, $$ f(H \times H) \subset H, \tag{0} $$ which is equivalent to saying that $$g(H \times H) \subset H, $$ because, for all $x \times y \in H \times H$, we have $$g(x \times y) = xy^{-1} = f(x \times y). $$
Let $h \colon H \times H \to H$ be the map obtained from $g$ by restricting the range of $g$ to the subset $H$ of $G$.
Now as the map $g \colon H \times H \to G$ is continuous and as $g( H \times H) \subset H$, so by Theorem 18.2 (e) in Munkres, the map $h$ defined in the preceding paragraph is continuous.
Thus $H$ is a group that is also a topological space satisfying the $T_1$ axiom and the map $h \colon H \times H \to H$, defined by $$ h( x \times y) \colon= xy^{-1} \qquad \mbox{ for all } x \times y \in H \times H,$$ is continuous. Hence $H$ is a topological group.
Is this part of my proof correct?
As $H \subset G$, so by Prob. 9, Sec. 17, in Munkres, we have the identity $$ \overline{H \times H} = \overline{H} \times \overline{H}. \tag{1} $$
And, as $H \times H \subset G \times G$ and as the map $f \colon G \times G \to G$ defined in (A) above is continuous, so by Theorem 18.1 (2) in Munkres,
$$ f\left( \overline{H \times H} \right) \subset \overline{ f(H \times H)}. \tag{2} $$Therefore, we have $$ \begin{align} f \left( \overline{H} \times \overline{H} \right) &= f\left( \overline{H \times H} \right) \qquad \mbox{[ by virtue of (1) above ] } \\ &\subset \overline{ f(H \times H)} \qquad \mbox{ [ by (2) above ] } \\ &\subset \overline{H}. \qquad \mbox{ [ because $f(H \times H ) \subset H$, by (0) above ] } \end{align} $$
Now as $$ f \left( \overline{H} \times \overline{H} \right) \subset \overline{H}, \tag{3} $$ so using the definition of $f$ from (A) above we can conclude that, for any elements $x, y \in \overline{H}$, the element $xy^{-1}$ is also in $\overline{H}$. Thus $\overline{H}$ is a subgroup of $G$.
Using the same logic as has been employed for $H$, we can conclude that the (topological) subspace $\overline{H}$ of $G$ is also a topological space satisfying the $T_1$ axiom.
As $\overline{H} \subset G$, so $\overline{H} \times \overline{H} \subset G \times G$ also, again by Prob. 2 (j), Sec. 1, in Munkres.
Now as the map $f \colon G \times G \to G$ defined in (A) above is continuous, so is the map $\tilde{g} \colon \overline{H} \times \overline{H} \to G$ obtained from $f$ by restricting the domain to the subset $\overline{H} \times \overline{H}$ of $G \times G$, by Theorem 18. 2 (d) in Munkres.
Moreover, from (3) above, we also obtain $$ \tilde{g} \left( \overline{H} \times \overline{H} \right) = f \left( \overline{H} \times \overline{H} \right) \subset \overline{H}. $$ Thus Theorem 18.2 (e) in Munkres guarantees the continuity of the map $\tilde{h} \colon \overline{H} \times \overline{H} \to \overline{H}$ obtained from $\tilde{g}$ by restricting the range to the subset $\overline{H}$ of $G$.
But the formula for the map $\tilde{h}$ is as follows. $$ \tilde{h} ( x \times y) = g( x \times y) = f(x \times y) = xy^{-1} \qquad \mbox{ for all } x \times y \in \overline{H} \times \overline{H}. $$
Thus we have shown that $\overline{H}$ satisfies all the requirements for being a topological group.
Is this part of my proof correct?
If both parts of my proof are correct, is my presentation also accessible enough for somebody studying this stuff for the very first time?
Logically, I found only one shortcoming: You use the fact that $H \times H$ is a subspace (not just subset) of $G \times G$ implicitly, without acknowledgement or proof (the reference you gave says only that $H \times H$ is a subset of $G \times G$). The issue is, the topology on $H \times H$ is the product topology obtained from the subspace topology on $H$. However, as a subspace of $G \times G$, the topology is the subspace topology inherited from the product topology on $G \times G$. Your map $h : H \times H \to H$ has to be continuous with respect to the product-of-subspaces topology in order for $H$ to be a topological group. But you prove its continuity from the restriction of $f$ from $G \times G$ to the subspace-of-product topology on $H \times H$. So it is important that these two topologies are the same.
Other than this incompleteness, both proofs appear to be logically correct. However, they are very difficult to follow because you belabor trivial matters far too exhaustively. This gets in the way of being able to follow your logic.
Some things you can do to improve readability: