Help grasping intuitively: if $\{p_n\}$ is a sequence in a compact metric space $X$, then some subsequence of $\{p_n\}$ converges to a point in $X$ This is theorem 3.6 in Rudin's Principles of Mathematical Analysis. I actually followed the logical steps of the proof, but I can't really picture why it's true like I can with some of these other theorems. Intuitively I feel like it shouldn't be so hard to construct a counterexample. I'm imagining that $X$ is, say, a closed ball in $\mathbb{R}^2$, and then we have a sequence which follows some weird spirally path all over and around the inside of $X$—think a spirograph where the radius of the inner circle varies randomly or something. I can't imagine how it is that such a sequence would have a convergent subsequence. Is there something wrong with my example? Or am I just not seeing something that I ought to be? 
 A: It might be a good idea to really write down an example of the kind you describe,
and see how it necessarily contains a convergent subsequence.
So let's suppose that our compact subset is the closed disk (ball) in $\mathbb R^2$, which I'll identify with $\mathbb C$ (so that it's easy to use polar coordinates), and let our sequence be $r_n e^{i \theta_n}$.  In your example, you imagine that the $r_n$ are varying like crazy, and so are the the $\theta_n$s.
Still, we will find a convergent subsequence!
First, let's focus on the $r_n$.  They are a sequence in the interval $[0,1]$.  Now the bisection proof that litteO mentions in a comment applies.  That is, cut $[0,1]$ into the two intervals $[0,1/2]$ and $[1/2, 1]$.  At least one of these contains infinitely many of the $r_n$.  Continuing with such bisections, we find a subsequence $r_{n_i}$ that converges to some $r \in [0,1]$.   Let's replace our sequence $r_n e^{i \theta_n}$ by the subsequence $r_{i_n} e^{i\theta_{n_i}}$, and then relabel, so that we may assume in our original sequence $r_n e^{i \theta_n}$ that the radii $r_n$ are actually converging to some fixed radius $r$.
Now if the limiting radius if $r = 0$, then in fact our sequence $r_n e^{i\theta_n}$ is converging to $0$, and we're done.  If $r \neq 0,$ then 
what we know is that the points $r_n e^{i \theta_n}$ get closer and closer to the circle of radius $r$, but we don't have any control of their angles $\theta_n$.  But now we can focus on the $\theta_n$ (forgetting the $r_n$ for a moment), and making the same bisection argument, on the interval $[0,2\pi]$ now,
we can find a subsequence of angles $\theta_{n_i}$ which converge to a limit, say $\theta$,
and so passing to this subsequence, we find that it converges to $r e^{i\theta}$.
Conclusion: no matter how crazily $r_n$ and $\theta_n$ vary, the bisection argument shows that some subsequence of the $r_n$ has to tend towards a limit, and then for some subsequence of this subsequence, the angles also have to tend towards a limit (unless the subsequence of $r_n$ tends towards $0$, in which
case we don't have to worry about the angles at all).
The bisection argument shows that as long as we are looking at quanities in a bounded closed interval, there is just "not enough space" to avoid the existence of a convergent subsequence, however much the original sequence is varying.
Additional remark: Of course, one can prove the result in the case of a disk just by embedding it into a square, and putting a finer and finer mesh over the square (the two-dimensional version of the bisection argument).  I phrased the above argument in terms of the polar coordinates because it seemed to fit with the kind of counterexample you were trying to imagine.
A: In $\mathbb R$ (or $\mathbb R^n$) you can give an intuitive bisection-style proof.
Let $I = [a,b] \subset \mathbb R$ be a closed interval and let $(p_n)_{n=1}^{\infty}$ be a sequence of numbers in $I$.  Let $L_1 = [a,\frac{a+b}{2}]$ and $R_1 = [\frac{a+b}{2},b]$.  If $L_1$ contains infinitely many points of $(p_n)_{n=1}^{\infty}$, then let $I_1 = L_1$.  Otherwise, let $I_1 = R_1$.  So $I_1$ contains infinitely many points of $(p_n)_{n=1}^{\infty}$.
Now subdivide $I_1$ in the same way, and continue in this manner, creating an infinite nested sequence of closed intervals $I \supset I_1 \supset I_2 \supset \cdots$.  The intersection $\cap_{i=1}^{\infty}$ is non-empty and contains a single point $p$.  You can see that there is a subsequence of $(p_n)_{n=1}^{\infty}$ which converges to $p$.
