In an answer to another question, it is remarked that there are fibre sequences $$\mathbf{RP}^2 \to BO(2) \to BSO(3),$$ $$\mathbf{RP}^\infty \to BSU(2) \to BSO(3).$$
The second one is explained by another fact stated in this answer, namely: if we have an exact sequence $1 \to H \to G \to G/H \to 1$, then it induces a fibre sequence $$ BH \to BG \to B(G/H).$$ I don't know how to obtain the first one though.
Does someone have a reference for either of these facts?
You can directly verify that a short exact sequence of groups induces a fiber sequence of classifying spaces by using the long exact sequence in homotopy to compute say the homotopy groups of the fiber, and noting that classifying spaces are characterized by their homotopy groups.
Or you can write down models of classifying spaces and check. Let $EG$ be a contractible space on which $G$ acts freely, etc. Then the map $BG \to B(G/H)$ is modeled by $$BG \simeq EG \times_G E(G/H) \simeq BH \times_{G/H} E(G/H) \to * \times_{G/H} E(G/H) \simeq B(G/H),$$ whose homotopy fiber is $BH$.
On the other hand, the sequence $$\mathbb{R}P^2 \to BO(2) \to BSO(3)$$ is induced by the group homomorphism $O(2) \hookrightarrow SO(3)$ sending $A$ to $A \oplus \det A$. The fiber of $BO(2) \to BSO(3)$ is $SO(3)/O(2)$, which can be identified with $\mathbb{R}P^2$. ($SO(3)$ acts transitively on lines in $\mathbb{R}^3$, and the stabilizer of a line is $O(2)$.)
