An intuitive idea about fundamental group of $\mathbb{RP}^2$

The following argument is essentially an application of the path lifting property for covering spaces.

Let's think about $\mathbb{R}P^2$ as being the quotient space you get by identifying antipodal points on the sphere $S^2$. That is, let $x\sim -x$, let $\mathbb{R}P^2=S^2/\sim$ and let $p\colon S^2\rightarrow\mathbb{R}P^2$ be the quotient map. Let $z$ be the base point of $S^2$ and $y$ be the base point of $\mathbb{R}P^2$.

Now, consider a non-trvial loop $\gamma\colon[0,1]\rightarrow\mathbb{R}P^2$ based at the point $y\in\mathbb{R}P^2$ (so $\gamma$ can not be homotoped to a constant loop). Note that the preimage of $y$ under $p$ is exactly two points in $S^2$ which are $z$ and $-z$. If we lift the loop $\gamma$ up to $S^2$ via the lift $\tilde{p}$, the end points of the lifted path $\tilde{\gamma}\colon[0,1]\rightarrow S^2$ will either both be at $z$, or $\tilde{\gamma}(0)=z$ and $\tilde{\gamma}(1)=-z$.

But note that if both end points are at $z$, then $\tilde{\gamma}$ is a loop and we know that $S^2$ is simply connected so such a loop can be homotoped to a constant loop. Such a homotopy induces a similar homotopy in the loop $\gamma$ and so $\gamma$ must be trivial. This is a contradiction as we asked for $\gamma$ to be non-trivial. So, $\tilde{\gamma}(0)=z$ and $\tilde{\gamma}(1)=-z$.

Now, in this case, the path $\tilde{\gamma}$ can not be homotoped to a constant loop without moving the fixed ends of the path but if we consider the lift of the path $2\gamma$ via $\tilde{p}$, then the lifted path $\tilde{2\gamma}$ is a loop in $S^2$. Again, $S^2$ is simply connected and so such a loop can be homotoped to a constant loop and such a homotopy induces a similar homotopy in the loop $2\gamma$ and so $2\gamma$ is a trivial loop.

Try watching Your palm is a spinor on youtube. This move is part of a traditional phillipine dance – watch about 40 seconds into the clip.

As you go from the performer's more or less stationary shoulder to the hand that holds the glass, you are in fact following a homotopy from the trivial loop to the loop that rotates 720 degrees around a vertical axis.

The move is not hard to learn. But try it with an empty glass at first.

Edit: I neglected to add that this is really about $\mathrm{SO}(3)\simeq\mathbb{R}P^3$, not $\mathbb{R}P^2$. It's the same sort of thing going on, really. To see that $\mathrm{SO}(3)\simeq\mathbb{R}P^3$, think of a rotation as specified by a vector $x\in\mathbb{R}$ with $\|x\|\le\pi$, the direction giving the axis and the length the angle of rotation in the positive direction, as seen from the positive end of the axis. This identifies antipodal points on the sphere of radius $\pi$, thus turning the closed ball into a projective 3-space.

Here is a bit more algebraic perspective: if $X$ and $Y$ are path connected and $p: Y\rightarrow X$ is a covering map, then the number of sheets is equal to index of the subgroup $p_*(\pi_1(Y))$ in $\pi_1(X)$. Since $S^2 \rightarrow \mathbb{R}P^2$ is a 2-sheeted universal cover, it follows that $\pi_1(\mathbb{R}P^2)$ has 2 elements (trivial subgroup has index 2).