In any case, the proof is very simple. Consider the $3$-form $$ \phi_0 = dx^1\wedge dx^2\wedge dx^3 + dx^4\wedge dx^5\wedge dx^6. $$ I claim that the subgroup $G\subset\mathrm{GL}(6,\mathbb{R})$ that stabilizes $\phi_0$ consists of the obvious subgroup $G_0=\mathrm{SL}(3,\mathbb{R})\times\mathrm{SL}(3,\mathbb{R})$ that fixes the two summands plus the discrete part that switches the two summands. Since this subgroup has dimension $8{+}8=16$ while $\mathrm{GL}(6,\mathbb{R})$ has dimension $36$, it follows that the $\mathrm{GL}(6,\mathbb{R})$-orbit of $\phi_0$ has dimension $20= 36-16$, which is the dimension of $\Lambda^3(\mathbb{R}^6)$. Thus, this orbit is open.
To see the claim, just look at the vectors $v\in \mathbb{R}^6$ that satisfy $\bigl(\iota_v(\phi_0)\bigr)^2=0$. Clearly, such a vector must lie in either the 'first' $\mathbb{R}^3$ or the 'second' $\mathbb{R}^3$. Thus, $G$ must carry each of these subspaces into either itself or the other subspace. The subgroup of $G$ that fixes each subspace must be of index $2$ in $G$, and is clearly $G_0$.
What is not so obvious is that there exist exactly $2$ open orbits. One is the one listed above, and the other is $$ \phi_1 = \mathrm{Re}\bigl((dx^1+i\ dx^4)\wedge(dx^2+i\ dx^5)\wedge(dx^3+i\ dx^6)\bigr), $$ whose stabilizer has an index two subgroup that is isomorphic to $\mathrm{SL}(3,\mathbb{C})$.
Remark: By the way, the existence of the open orbit shouldn't be surprising. It's a general fact that $\mathrm{GL}(n,\mathbb{R})$ has an open orbit in $\Lambda^k(\mathbb{R}^n)$ for $k>0$ whenever $n^2$, the dimension of $\mathrm{GL}(n,\mathbb{R})$, is as large as $n\choose k$, the dimension of $\Lambda^k(\mathbb{R}^n)$. This happens for all $n$ when $k\in\lbrace1,2,n{-}2,n{-}1,n\rbrace$ and otherwise only for $$ (n,k)\in\lbrace (6,3), (7,3), (7,4), (8,3), (8,5)\rbrace. $$