fibration of real projective space over complex projective space From the fibration $$U(1)=S^1\to S^{2n+1}\to \mathbb{C}P^n,$$ can we quotient the action of $S^0$ and obtain a well-defined fibration
$$
S^1/S^0=\mathbb{R}P^1\cong S^1\to \mathbb{R}P^{2n+1}\to\mathbb{C}P^n?$$
Consequently, can we obtain a fibration of limit
$$
S^1\to \mathbb{R}P^{\infty}\to\mathbb{C}P^\infty?
$$
 A: Yes, this is true and to put it a broader context, it can be seen as a special example of a more general principle:
Suppose we have a topological group $G$ and a subgroup $H$, then we get a fibre bundle $$G/H\rightarrow BH\rightarrow BG.$$ One way to construct this bundle is to start with a model of the universal principal $G$-bundle $$G\rightarrow EG\rightarrow BG$$ and mod out the action of $H$ to obtain a fibre bundle $$G/H\rightarrow EG/H\rightarrow BG$$ and realizing that $EG/H$ is a model for $BH$ since we have a principal $H$ bundle bunde $$H\rightarrow EG\rightarrow EG/H.$$
To specialise to your situation: $\mathbb{C}P^{\infty}$ is a model for $BU(1)=BS^1$ and $\mathbb{R}P^{\infty}$ is one for $BO(1)=B\mathbb{Z}/2\mathbb{Z}$ since we have principal fibre bundles $$U(1)\rightarrow S^\infty\rightarrow \mathbb{C}P^{\infty}$$ and $$O(1)\rightarrow S^\infty\rightarrow \mathbb{R}P^{\infty}$$ obtained as the colimit of the principal bundles $$U(1)\rightarrow S^{2n+1}\rightarrow \mathbb{C}P^{n}$$ and $$O(1)\rightarrow S^{n}\rightarrow \mathbb{R}P^{n}.$$ These are univeral since $S^{\infty}$ is contractible. 
The fibre bundle $$U(1)/O(1)=S^1\rightarrow BO(1)=\mathbb{R}P^{\infty}\rightarrow BU(1)=\mathbb{C}P^{\infty}$$ you get out of this construction is exactly the one you mentioned.
