How do I prove this: If $\mathcal{O}(p,q)$ is a Lie algebra of the semiorthogonal group $O(p,q)$ then $\mathcal O(p,q)$ consist of all matrices of the form:
$$X= \left( \begin{matrix} a & x^t\\ x & b \\ \end{matrix} \right),$$
where $a\in \mathcal O(p)$ and $b\in \mathcal O(q)$, and $x$ is an arbitrary real matrix $q \times p$.
Well you just do the same thing we do every night, Pinky, try to take over the world! Hm, well, I mean, compute differentials of curves.
Suppose that $\gamma$ is a smooth curve in $O(p,q)$ such that $\gamma(0)=I$, the identity matrix. If $x$ and $y$ are two vectors and $\def\i#1{\langle#1\rangle}$ is the usual quadratic form of signature $(p,q)$ on $\mathbb R^{p+q}$ then $\i{\gamma(t)x,\gamma(t)y}$ does not depend on $t$, so differentiating with respect to $t$ we see that $$0=\frac{\mathrm d}{\mathrm dt}\i{\gamma(t)x,\gamma(t)y}=\i{\gamma'(t)x,\gamma(t)y}+\i{\gamma(t)x,\gamma(t)'y}$$ and evaluating at $t=0$, $$0=\i{\gamma'(0)x,y}+\i{x,\gamma(0)'y}.$$ If $x=e_i$ and $y=e_j$ are vectors in the standard basis, then this tells us that $$a_{i,j}\epsilon_j+a_{j,i}\epsilon_i=0.$$ Here $a_{i,j}$ is the $(i,j)$th entry in the matrix $\gamma'(0)$ and $\epsilon_i$ is $1$ if $1\leq i\leq p$, and $-1$ if $p+1\leq i\leq p+q$. These equations imply that the matrix $\gamma'(0)$ is of the form you want.
This means that the tangent vector space at $I$ is contained in the vector space you wrote in your question. Now compare dimensions.
If you do not know already the dimension of the group, well, you need more work. Let now $A$ be a matrix in the space you wrote and consider the curve $\gamma:t\mapsto e^{At}$. More or less the same reasoning as above show that the function $t\mapsto\i{\gamma(t)x,\gamma(t)y}$ is constant, so that the curve is contained in your group. As its tangent vector at zero is the matrix $A$, this proves the other containment.