The orthogonal group is compact (this does not depend on whether we are looking at the subspace topology on $\textrm{O}_n$). Now firstly it is a closed subset of $M_n(\Bbb{R})$ - to prove this note that the map $f : M_n(\Bbb{R}) \to M_n(\Bbb{R})$ that sends a matrix $A$ to $AA^T - I$ is continuous with kernel precisely $\textrm{O}_n$. Furthermore the columns of an orthogonal matrix being orthonormal imposes the condition that there is a constant $C$ such that $|a_{ij}|\leq C$ for all $1 \leq i,j \leq n$. The result you are asking for now follows because each orthogonal matrix has an open neighbourhood (namely the whole of $\textrm{O}_n$) about it that is compact.