Suppose that $H \leq G$ , $\phi \in Char(H)$ and $K \leq G$ that $(\phi^G)_K \in Irr(K)$. We want to prove that $G=HK$. Suppose that $H \leq G$ , $\phi \in Char(H)$ and $K \leq G$ that $(\phi^G)_K \in Irr(K)$. We want to prove that $G=HK$.
I can prove it by modules but can anybody help to prove it without using modules.
 A: Edit: Adding a solution based on the hint in the comments.
We first use Frobenius reciprocity heavily. First we see that
$$
[\phi,(\phi_{H\cap K})^H]_H=[\phi_{H\cap K},\phi_{H\cap K}]_{H\cap K}>0.
$$
So $\phi$ shares an irreducible summand with $(\phi_{H\cap K})^H$. Therefore
$\phi^G$ also shares at least one summand with $(\phi_{H\cap K})^G$ (transitivity of induction). Two more applications of Frobenius reciprocity (and transitivity of restriction) then imply that
$$
[(\phi^G)_K,(\phi_{H\cap K})^K]_K=[(\phi^G)_{K\cap H},\phi_{H\cap K}]_{H\cap K}=
[\phi^G,(\phi_{H\cap K})^G]_G>0.
$$
Here it was assumed that $M=(\phi^G)_K$ is irreducible, so it has to be a summand of $N=\phi_{H\cap K}^K.$ But $\dim M= [G:H]\dim\phi$, and
$\dim N=[K:K\cap H]\dim\phi$, so we can conclude that $[K:K\cap H]\ge [G:H]$.
Consider two cosets $Hk$ and $Hk'$ with $k,k'\in K$. They are equal if and only if $k'\in Hk$ which happens if and only if $k'\in(H\cap K)k$. Therefore the number of distinct cosets of the form $Hk, k\in K,$ equals $[K:H\cap K]$.
But there can be at most $[G:H]$ such cosets, and we just saw that $[G:H]\le [K:K\cap H]$. Therefore all the cosets of $H$ in $G$ are in $HK$ and the claim follows.
Note that we also get as a corollary that $[G:H]=[K:H\cap K]$. Counting dimensions then implies that we actually have
$$
(\phi^G)_K=(\phi_{H\cap K})^K,
$$
as the l.h.s. is an irreducible summand of the r.h.s. and their dimensions are equal.

[My first answer based on Mackey's theorem below. Swat them flies with cannonballs!]
I'm not sure that this is what you want to see, but this follows immediately from Mackey's theorem: If $T$ is a set of double coset representative, i.e. $G$ is the disjoint union of the sets $HtK, t\in T$, then for all characters $\phi$ of $H$ we have
$$
(\phi^G)_K=\sum_{t\in T}(\phi^t_{H^t\cap K})^K.
$$
Here the superscript $t$ indicates conjugation by $t$, so $\phi^t$ is a character of $H^t$. 
As $(\phi^G)_K$ is irreducible, there can be only a single summand. Hence $|T|=1$, and we are done. As a bonus we get that
$$
(\phi^G)_K=(\phi_{H\cap K})^K.
$$
