Take $U$ a finite dimensional representation of $G\times H$. Consider only the $G$ action. Then every element from $H$ is an intertwining operator. Decompose $U$ into $G$-isotypic components. Then $H$ invariates each of them. Note that for every irreducible representation of $G$ the intertwining operators are scalers. Therefore, each $G$ isotypic of type $\rho$ of $U$ can be written as $\rho\otimes \eta$, where $\eta$ is a representation of $H$. Therefore, each finite dimensional (continuous) representation of $G\times H$ can be written as a direct sum of $\rho_1\otimes \rho_2$, where $\rho_1$ is an irrep of $G$ and $\rho_2$ is an irrep of $H$. From here it is easy also to show that $\rho_1\otimes \rho_2$ are irreducible.
This argument avoids most of the analysis. It only uses Schur's lemma on intertwining operators. One needs to argue that $\eta$ above is continuous, if we started with a continuous representation of $G\times H$, which is straightforward, looking at matrix coefficients.
There are some important represantation theoric properties which are lost when $F$ is not algebraically closed.
Lemma $1$: Let $A$ be a finite dimensional $F$-algebra where $F$ is algebraicly closed and $V$ be an irreducable $A$-module. Then $End_A(V)\cong F$.
Lemma $2$: Let $A$ be a finite dimensional commutative $F$algebra where where $F$ is algebraically closed and $V$ be an irreducible $A$-module. Then $dim_{ F}(V)=1$.
Above lemma says that if $G$ is abelian and $V$ is an $FG$ module where $F$ is algebraicly closed then $dim_{ F}(V)=1$. On the other hand, if $F=\mathbb{R}$ then we can only say that $dim_{ \mathbb R}(V)\leq 2.$
Thus, number of the linear characters of $G$ need not to be equal $|G|$ in that case.
(Notice that when $G$ is abelian number of the conjugacy classes is $|G|$).
Notice that Mackey's theorem works for fields whose characteristic is zero. Thus the group algebra $FG$ is completely reducible when $F=\mathbb R$ or $F=\mathbb C$. From now on assume characteristic of $F$ is zero.
Now let $V$ be an irreducible $A=FG$ (right) module and $0\neq v\in V$. Since $vA$ is $A$-invariant space, we get $V=vA$ due to the simplicity of $V$. Then $V\cong A/ann(v)$. Since $A$ is completly reducable $A$-modue, $A=Ann(v)\oplus U$ where $U$ is $A$-submodue of $A$. Then we see that $U$ is isomorphic to $V$, that is, every irreducible $A$-module is isomorphic to a submodule of $A$ (this also show that there are finitely many distinct irreducible $A$-module).
Now suppose that $F=\mathbb C$. There are many ways of showing that the number of the irreducable character of $G$ (equivalently, the number of the irreducable $A$-module up to isomorphism) is equal to the number of conjugacy classes of $G$.
Each author follows a distinct way. Some of them defines the concept of "characters" and "class function" then shows that the set of irreducable characters constiyes a ortanormal base for the spaces of all class functions which has a dimension equal to the number of the conjugacy classes of $G$. (You can read the chapter 15 of "Algebra: A Graduate Course - I. Martin Isaacs". The the theorem you are looking for is Theorem 15.5) Another approach is purely by "represantation theory".
Now assume $F=\mathbb R$ and $G$ is abelian so that $FG$ is a commutative algebra. Since Lemma $2$ is no longer true, the number of the irreducable character of $A$ could be less than the dimension of $A$, which is eqaul to $|G|$, as some of the irreducable $A$ moduel can be of dimension $2$.
Best Answer
You should note that the dimension of the space of class functions is equal to the number of conjugacy classes, and that orthonormal vectors in a Hermitian inner product space are linearly independent.