Is $O_n({\bf Q})$ dense in $O_n({\bf R})$?

There's an easy argument based on the Cayley transform: If $a$ is a skew-symmetric $n$-by-$n$ real matrix, then $I_n+a$ is invertible (since $(I_n-a)(I_n+a)=I_n-a^2$ is a positive definite symmetric matrix and hence invertible), and $$ A = (I_n-a)(I_n+a)^{-1} $$ is orthgonal (i.e., $AA^T = I_n$). Note that $(I_n+A)(I_n+a) = 2I_n$, so $I_n+A$ is invertible. Conversely, if $A$ is an orthogonal $n$-by-$n$ matrix such that $I_n+A$ is invertible, one can solve the above equation uniquely in the form $$ a = (I_n+A)^{-1}(I_n-A) = -a^T. $$ This establishes a rational 'parametrization' (known as the Cayley transform) of $\mathrm{SO}_n(\mathbb{R})$. Plainly, $a$ has rational entries if and only if $A$ has rational entries.

The density of $\mathrm{O}_n(\mathbb{Q})$ in $\mathrm{O}_n(\mathbb{R})$ follows immediately.

By Cartan-Dieudonné's theorem, every element of $O_n({\bf R})$, resp. $O_n({\bf Q})$ is a product of at most n hyperplane reflections $\sigma_u$ for u in ${\bf R}^n$, resp. u in ${\bf Q}^n$. Now it suffices to remark that a reflection is a limit of rational reflections.

Yes, here'a a proof by induction, granted the $n=2$ case (which is the only one where [basic] arithmetic occurs).

Let $G$ be the closure. I first claim that $G$ acts transitively on the sphere. Indeed, let $x=(x_1,\dots,x_n)$ be on the sphere. Using the case $n=2$ on the last two coordinates, we see that $x$ is in the orbit of some $y=(y_1,\dots,y_n)$ with $y_n=0$. Using the case in dimension $n-1$, we deduce that $y$ is in the orbit of $e_1=(1,0,\dots,0)$.

Now let $g$ be in $\mathrm{O}(n)$. By the claim, there exists $h\in G$ such that $g(e_1)=h(e_1)$. So $g^{-1}h$ fixes $e_1$, hence belongs to the copy of $\mathrm{O}(n-1)$ acting on the last $n-1$ coordinates. By induction, $g^{-1}h\in G$. So $g\in G$.