Visualizing $|\mathcal{B}\mathbb{Z}| \simeq S^1$. The classifying space of the integer group $\mathbb{Z}$ can be defined as the geometric realization of the underlying groupoid $\mathcal{B}\mathbb{Z}$. 
To unwind, $\mathcal{B}\mathbb{Z}$ is simply the category with one object, with $\mathbb{Z}$ as its morphism space. The geometric realization $|\mathcal{B} \mathbb{Z}|$ is a topological space (CW-complex) defined inductively:


*

*To each object, assign a point.

*To each morphism, assign a 1-disk (segment) with corresponding end points.

*To each 2-tuple of composable morphisms $(f,g)$, assign a 2-disk with corresponding 1-disks as boundary.

*To each 3-tuple of composable morphisms $(f,g,h)$, assign a 3-disk... and so on.


It seems like a huge topological space, but there's a theorem I heard several times stating that it's really the classifying space of principal $\mathbb{Z}$-bundles, which turns out to be the $1$-sphere $S^1$ homotopically.
My ultimate goal is to understand (theoretically and pictorially) the proof of the general statement above, for general groups $G$ instead of just $\mathbb{Z}$. I think it will be good to start with this simplest case.
Question: However, I find it rough to visualize how the infinitely defined space above is homotopic to $S^1$. Could you point out how?
More fun: think in this vein for that $|\mathcal{B}\mathbb{Z_2}|$ is $RP^\infty$ and that $|\mathcal{J}|$ is $E\mathbb{Z}_2$, where $\mathcal{J}$ is the unique category with two objects $X,Y$ and four morphisms $f:X\to Y, g:Y\to X$.
 A: Since the homotopy type of spaces is determined by their homotopy groups, at least on a convenient category of spaces, it suffices to show that the homotopy groups of $\mathcal{B}\mathbb{Z}$ coincide with those of $S^1$, that is:
$$
\pi_n(\mathcal{B}\mathbb{Z}) = 
\begin{cases}
\mathbb{Z}
&&\text{ if }
n = 1
\\
0
&&\text { if }
n\neq 1
\end{cases}
$$
This is equivalent to showing that $\mathcal{B}\mathbb{Z}$ is the first Eilenberg-MacLane space $K(1,\mathbb{Z})$ are weakly homotopic, and in fact this is true for any discrete group $G$.
After a couple well known lemmas, the proof outlined below becomes elementary.
First notice that for a discrete group $G$ we have
$$
\pi_n(G) = 
\begin{cases}
G
&&\text{ if }
n = 0
\\
0
&&\text { if }
n > 0
\end{cases}.
$$
Recall that $[S^n,X]\cong [S^{n-1},\Omega X]$, and that $\mathcal{B}$ is a delooping i.e. $\Omega\mathcal{B} G\cong G$.
Proposition: If $G$ is discrete then $\mathcal{B}G$ is the first Eilenberg-Maclne $K(1,\mathbb{Z})$.
Proof:
For $n =1$,
\begin{align*}
\pi_1(\mathcal{B}G) 
&=
[S^1,\mathcal{B}G]
\\
&\cong
[S^0,\Omega \mathcal{B}G]
\\
&\cong
[S^0,G]
\\
&\cong
\pi_0(G)= G.
\end{align*}
For $n\geq 2$,
\begin{align*}
\pi_n(\mathcal{B}G) 
&=
[S^n,\mathcal{B}G] 
\\
&\cong
[S^{n-1},\Omega \mathcal{B}G]
\\
&\cong
[S^{n-1},G]
\\
&\cong
\pi_{i-1}(G) = 0.
\\
\end{align*}
$$
\hspace{15cm}\blacksquare
$$
It is interesting to think of what could happen if $\mathcal{B}$ could be iterated i.e. $\mathcal{B}^nG$ made sense.
The condition is that $\mathcal{B}G$ is itself a group; surprisingly, for abelian $G$ that can be done and $\mathcal{B}G$ is abelian.
For abelian and discrete $G$, such as $\mathbb{Z}$, one could reason exactly as above and show that $K(n,G)\cong \mathcal{B}^n G$.
Particularly, we have $\mathcal{B}^n \mathbb{Z}\cong S^n$.
For your More fun question, notice that the question now boils down to showing that $\pi_1(\mathbb{RP}^\infty) = \mathbb{Z}_2$ and all higher homotopy groups vanish.
That follows from its orientable double cover $S^\infty$ being contractible.
I have written some account of these facts on a set of lecture notes written for a short course I taught earlier this year; see Section 3.
You can also find a lot insight at this blog post by Baez.
