In the case of a fibre bundle with a discrete fibre, in particular as given in the example below,
The covering map $S^n \rightarrow \mathbb{R} \mathbb{P}^n$ gives a fibre bundle $\{\pm 1\} \hookrightarrow S^n \rightarrow \mathbb{R} \mathbb{P}^n$. Note that, when $n=1$, this specialises to the sphere bundle $S^0 \hookrightarrow S^1 \rightarrow S^1$.
I wonder how is it that an $S^0$(two disjoint points) fibre produce no change to the base space? What I imagine is that, an $S^0$ fibre, must create two concentric $S^1$s under the inverse projection map, and I think that is topologically different from a single circle(though I am not sure about it). where am I getting wrong?
Note that an $S^0$-bundle is the same thing as a double cover (since $S^0$ has two points). While it is true that $S^0 \to S^1\times\{0,1\} \xrightarrow{\pi} S^1$ with $\pi(z, i) = z$ is a double cover of $S^1$, that is not the double cover that is being referenced. Instead, it is the connected double cover of $S^1$ given by $S^0 \to S^1 \xrightarrow{\pi} S^1$ where $\pi(z) = z^2$ (thinking of $S^1$ as the unit circle in $\mathbb{C}$).
More generally, for any positive integer $n$, there is a degree $n$ covering map $\pi : S^1 \to S^1$ given by $\pi(z) = z^n$.