Here is Prob. 1 in the Supplementary Exercises in Chapter 2 in the book Topology by James R. Munkres, 2nd edition:
Let $H$ denote a group that is also a topological space satisfying the $T_1$ axiom. Show that $H$ is a topological group if and only if the map of $H \times H$ into $H$ sending $x \times y$ into $x \cdot y^{-1}$ is continuous.
And, here is Munkres' definition of topological group:
A topological group $G$ is a group that is also a topological space satisfying the $T_1$ axiom, such that the map of $G \times G$ into $G$ sending $x \times y$ into $x \cdot y$, and the map of $G$ into $G$ sending $x$ into $x^{-1}$, are continuous maps.
My Attempt:
Suppose first that $H$ is a topological group. Then the maps $f \colon H \times H \to H$ and $g \colon H \to H$, defined by $$ f(x \times y) \colon= x y \qquad \mbox{ for all } \ x \times y \in H \times H $$ and $$ g(x) \colon= x^{-1} \qquad \mbox{ for all } \ x \in H,$$ are continuous.
And, of course the identity map $i_H \colon H \to H$ defined by $$ i_H (x) \colon= x \qquad \mbox{ for all } \ x \in H,$$ is also continuous.
Now as $i_H \colon H \to H$ and $g \colon H \to H$ are continuous, so is the map $i_H \times g \colon H \times H \to H \times H$ defined by $$ \left( i_H \times g \right) (x \times y) \colon= i_H(x) \times g(y) = x \times y^{-1} \qquad \mbox{ for all } \ x \times y \in H \times H,$$ by virtue of Prob. 10, Sec. 18, in Munkres.
Finally, as the maps $i_H \times g \colon H \times H \to H \times H$ and $f \colon H \times H \to H$ are continuous, so is the composite map $f \circ \left( i_H \times g \right) \colon H \times H \to H$, by Theorem 18.2 (c) in Munkres. But, for each $x \times y \in H \times H$, we see that $$ \left( \ f \circ \left( i_H \times g \right) \ \right) (x \times y) = f \left( \ \left( i_H \times g \right) \ \right) (x \times y) = f \left( x \times y^{-1} \right) = xy^{-1}. $$ If we put $h \colon= f \circ \left( i_H \times g \right) $, then 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, as required.
Conversely, suppose that $H$ is a group that is also a topological space satisfying the $T_1$ axiom, such that 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. Then, for each $a \in H$, the map $g_a \colon H \to H$, defined by $$ g_a(x) \colon= h(a \times x) = a \times x^{-1} \ \qquad \mbox{ for all } \ x \in H, $$ is also continuous, by virtue of Prob. 11, Sec. 18, in Munkres. Let $e$ denote the identity element in $H$. Thus, in particular, the map $g (\colon= g_e) \colon H \to H$ defined by $$ g(x) \colon= e \times x^{-1} = x^{-1} \qquad \mbox{ for all } \ x \in H, \tag{1} $$ is continuous.
Now as the identity map $i_H \colon H \to H$ is continuous and as the map $g \colon H\to H$ of the preceding paragraph is continuous, so is the map $i_H \times g \colon H \times H \to H \times H$, defined by $$ \left( i_H \times g \right) (x \times y) \colon= i_H(x) \times g(y) = x \times y^{-1} \qquad \mbox{ for all } \ x \times y \in H \times H.$$
Now as the maps $i_H \times g \colon H \times H \to H \times H$ and $h \colon H \times H \to H$ are continuous, so is the composite map $h \circ \left( i_H \times g \right) \colon H \times H \to H$, by Theorem 18.2 (c) in Munkres. But, for each $x \times y \in H \times H$, we see that $$ \left( \ h \circ \left( i_H \times g \right) \ \right) (x \times y) = h \left( \ \left( i_H \times g \right) (x \times y) \ \right) = h \left( x \times y^{-1} \right) = x\left(y^{-1}\right)^{-1} = xy. $$ Thus map $f \colon H \times H \to H$, defined by $$ f(x \times y) \colon= xy \qquad \mbox{ for all } \ x\times y \in H \times H, \tag{2} $$ is continuous.
Thus $H$ is a group that is also a topological space satisfying the $T_1$ axiom, for which we have just shown that the maps $g \colon H \to H$ and $f \colon H \times H \to H$ given, respectively, by (1) and (2) above are continuous. Hence $H$ is a topological group.
Is this proof sound enough? If so, then is the presentation explicit enough? If not, then at what point is it in need of improvement?