Linear subspace closed under all special orthogonal matrices 
Let $n\in \mathbb N$ and $E$ be an $n$ dimensional vector space over $\mathbb R$.
Let $F$ be a linear subspace of $E$ such that $\forall f\in SO(E), f(F)\subset F$
Prove that $F=\{0\}$ or $F=E$

Although the result is quite intuitive, I haven't been able to write up a proof. I've noticed that the orthogonal complement is closed under $SO(E)$ as well.
I tried a proof by contradiction. If $\dim F\in\{1,\ldots,n-1\}$, one may consider some well-chosen reflection...

To prove the result, it will suffice to show that if $F$ contains some line$~L_0$ through the origin (in other words if $F\neq\{0\}$) then $F$ contains every line through the origin (in other words if $F=E$). Let $L_1$ be an arbitrary line through the origin; one needs to find of $g\in SO(E)$ such that $g(L_0)=L_1$; then the property of being $SO(E)$-stable will force $F$ to contain $L_1$, because it contains $L_0$.
Your idea of using reflections works. After choosing spanning unit vectors $v_0,v_1$ of $L_0,L_1$ respectively (and assuming $v_0\neq v_1$) one can apply the orthogonal reflection in the hyperplane orthogonal to $v_1-v_0$ to send $v_0$ to $v_1$, and therefore $L_0$ to $L_1$. The reflection is in $O(E)$ but not in $SO(E)$, but this can be remedied by composing with the orthogonal reflection in the hyperplane orthogonal to $v_1$, which fixes $L_1$. The composition of both reflections is your $g$.
Another approach is to use orthonormal bases, and notably the fact that for a given line one can always choose an orthonormal basis starting with a vector from this line.
After choosing one reference ordered orthonormal basis$\def\B{\mathcal B}~\B$, the elements $g\in SO(E)$ are in bijection the set of ordered orthonormal bases of$~E$ with the same orientation as$~\B$; the bijection is simply taking the image of$~\B$ under$~g$. Now choose $\B$ to start with a vector from$~L_0$, and choose another orthonormal basis to start with a vector from$~L_1$; one can arrange (by flipping a sign in the latter basis if necessary) that the two bases have the same orientation. The unique $g\in SO(E)$ sending$~\B$ to the second basis will have $g(L_0)=L_1$.
