Reducible reductive Lie subalgebras of so(p,q) Is it true that $S(O(p) \times O(q))$ is the only proper subgroup of $SO(p,q)$ of full rank acting on the natural representation $\mathbb{R}^{p+q}$ of $SO(p,q)$ that stabilizes a $p$-dimensional positive subspace and a $q$-dimensional negative subspace, and no other proper non-zero subspace?. More particularly, letting $H := SO(p) \times SO(q)$, may I know if there are any proper subgroups $K$ of $H$ such that rank $H$ = rank $K$ and $K$ ``behaves" as $H$ does?. 
 A: I answer the second question.
First I classify the connected $\mathbb{R}$-subgroups of full (absolute) rank of the compact $\mathbb{R}$-group $SO(p,\mathbb{R})$
that are $\mathbb{R}$-irreducible in the standard representation in $\mathbb{R}^p$.
The connected $\mathbb{R}$-subgroups of full rank of $SO(p,\mathbb{R})$ correspond to
connected $\mathbb{C}$-subgroups of full rank of $SO(p,\mathbb{C})$,
and correspond to Lie subalgebras of full rank in the Lie algebra $\mathfrak{so}(p,\mathbb{C})$.
Maximal Lie subalgebras of full rank of simple complex Lie algebras were classified by Dynkin in 1952.
For a modern reference see Onishchik and Vinberg (eds.), Lie Groups and Lie Algebras III,
Encyclopaedia of Mathematical Sciences, Vol. 41, Springer-Verlag,  Berlin, 1994. See Tables 5 and 6 in this book.
Let us assume for simplicity that $p\ge 7$. Then it follows from Tables 5 and 6 that, up to conjugation,
the only maximal subalgebras of full rank of $\mathfrak{so}(p,\mathbb{C})$ are of the form $\mathfrak{so}(k,\mathbb{C})\oplus\mathfrak{so}(p-k,\mathbb{C})$ and,
when $p$ is even, $\mathfrak{gl}(p/2,\mathbb{C})$.
The Lie subalgebra $\mathfrak{so}(k,\mathbb{C})\oplus\mathfrak{so}(p-k,\mathbb{C})$ corresponds to the compact subgroup $SO(k,\mathbb{R})\times SO(p-k,\mathbb{R})$, which is reducible.
The Lie subalgebra $\mathfrak{gl}(p/2,\mathbb{C})$ corresponds to the compact subgroup $U(p/2)$, which is irreducible (over $\mathbb{R}$, though not over $\mathbb{C}$).
Now we know all the maximal connected $\mathbb{R}$-subgroups of full rank of $SO(p,\mathbb{R})$ that are $\mathbb{R}$-irreducible: we have only $U(p/2)$.
What about non-maximal subgroups? They must be proper connected subgroups of full rank of $U(p/2)$.
Any proper connected subgroup of full rank of $U(p/2)$ is contained in a maximal connected subgroup of full rank of $U(p/2)$.
From Tables 5 and 6 we see that the only maximal connected subgroups of full rank of $U(p/2)$
are of the form $U(k)\times U(p/2-k)$. They are reducible over $\mathbb{R}$.
Thus there are no  non-maximal connected subgroups of full rank of $U(p/2)$ which are irreducible over $\mathbb{R}$.
Now if $H=H_1\times H_2$, where $H_1=SO(p,R)$ and $H_2=SO(q,\mathbb{R})$, and $K\subset H$ is a subgroup of full rank, then $K=K_1\times K_2$,
where $K_1\subset H_1$ and $K_2\subset H_2$ are subgroups of full ranks. 
If $K$ acts irreducibly over $\mathbb{R}$ in $\mathbb{R}^p$ and $\mathbb{R}^q$, then $K_1$ acts irreducibly over $\mathbb{R}$ in $\mathbb{R}^p$ and
$K_2$ acts irreducibly over $\mathbb{R}$ in $\mathbb{R}^q$.
We obtain that  $K=K_1\times K_2$, where $K_1$ is either $SO(p,\mathbb{R})$ or $U(p/2)$, and $K_2$ is either $SO(q,\mathbb{R})$ or $U(q/2)$.
