How to demonstrate $SO(3)$ is not simply connected? A quote from Wikipedia's article on the Rotation group: 

Consider the solid ball in $\mathbb{R}^3$ of
  radius $\pi$ [...].
  Given the above, for every point in
  this ball there is a rotation, with
  axis through the point and the origin,
  and rotation angle equal to the
  distance of the point from the origin.
  The identity rotation corresponds to
  the point at the center of the ball.
  Rotation through angles between $0$ and
  $-\pi$ correspond to the point on the same axis and distance from the origin but
  on the opposite side of the origin.
  The one remaining issue is that the
  two rotations through $\pi$ and through $-\pi$
  are the same. So we identify [...] antipodal points on the
  surface of the ball. After this
  identification, we arrive at a
  topological space homeomorphic to the
  rotation group.

So far, so good. This illustrates $SO(3)\cong \mathbb{RP}^3$.

These identifications illustrate that
  $SO(3)$ is connected but not simply
  connected. As to the latter, in the
  ball with antipodal surface points
  identified, consider the path running
  from the "north pole" straight through
  the center down to the south pole.
  This is a closed loop, since the north
  pole and the south pole are
  identified. This loop cannot be shrunk
  to a point, since no matter how you
  deform the loop, the start and end
  point have to remain antipodal, or
  else the loop will "break open".

I believe that $SO(3)$ is connected but the "intuitive argument" for $\pi_1(SO(3))\neq 0$ is not clear to me: The starting point at the "north pole" is a rotation of $\pi$ counterclockwise around the $z$ axis. This agrees with the "south pole", a rotation of $\pi$ clockwise around the $z$ axis. So the described path is a full $2\pi$ rotation counterclockwise around the $z$ axis, stating not in the identity position. Why isn't this homotopic to the trivial path? Antipodal points are identified, so what does "start and end point have to remain antipodal, or else the loop will "break open"" mean?
 A: You have two different proofs (a.k.a. the cup proof and the belt proof) right here 
http://www.youtube.com/watch?v=CYBqIRM8GiY
There are several different demonstrations of the second one (Harrison Brown linked to this video in a comment to an answer I submitted - the cup proof - in the question asking for "proofs w/o words").
A: A loop is homotopically trivial if it can be continuously deformed to the constant loop.  This means that at every step of the deformation (every "instant in time") you still have a loop.  There are two kinds of loops on the unit ball with antipodal identifications in the boundary: either it's also a loop in the ball (without identifications) or else it starts and ends at antipodal points. The example curve in the question is of the latter kind.  It seems intuitive that any continuous deformation of this curve which remains closed has to still connect antipodal points, since you cannot move the ends closer to each other -- which is what you'd have to do in order to get a contractible loop -- while keeping it a closed curve.
A: Don't think about $SO(3)$ to start with, think about the unit quaternions, $S^3\subset \mathbb{R}^4$ where multiplication is given by quaternion multiplication and inverses are given by "complex conjugation".  It might help to realize that the dot product in $\mathbb{R}^4$ is given by $q.p=Re(q\overline{p})$.  
The unit quaternions act on themselves by conjugation, and this action fixes the identity, inducing an action of the unit quaternions on the tangent space of $S^3$ at $1$ which is canonically isomorphic to $\mathbb{R}^3$. Call the unit quaternions $S^3$, call this action $ad:S^3\times \mathbb{R}^3\rightarrow \mathbb{R}^3$.  First prove that this action is as rigid rotations by proving it preserves the dot product on $\mathbb{R}^3$.  Next prove that the kernel of the map induced
by $ad$, $h:S^3\rightarrow SO(3)$ has kernel $\{\pm 1\}$.  Finally, use a dimension count to prove the map is onto.  From there you can conclude that $S^3$ is the universal cover of $SO(3)$ with group of deck transformations $\mathbb{Z}_2$.
I usually assign this as homework, though I set the kids up by giving this outline to fill in. 
Another fun picture of $SO(3)$ is given by the unit tangent bundle of $S^2$.  Notice that this can be described as
$$ T_1S^2=\{(\vec{u},\vec{v})\in \mathbb{R}^3\times \mathbb{R}^3|||\vec{u}||=||\vec{v}||=1, \ \vec{u}.\vec{v}=0\} $$
Notice that the matrix with columns $\vec{u},\vec{v},\vec{u}\times \vec{v}$ is in $SO(3)$. This map gives a diffeomorphism between $T_1S^2$ and $SO(3)$.  The projection map $p:T_1S^2\rightarrow S^2$ that sends the pair $(\vec{u},\vec{v})$ to $\vec{u}$ is a fibration, and this is the standard fibration that people use to analyze the homotopy groups of $SO(3)$.
