L = {w belongs to {0, 1} * ∣ w has at most three 0’s}. Why is this regular language?
As the other answers have said, the key is in the statement of the pumping lemma.

The key to understanding how to apply the pumping (and how not to as well) is paying close attention to the quantifiers ("for all" and "there exists") and the order they come in:


  The Pumping Lemma for Regular Languages
  
  If the language $L \subseteq \Sigma^{\ast}$ is regular, then there exists a constant $p \geq 1$ such that for every string $s \in L$ where $|s| \geq p$, there exists a division of $s = xyz$ such that:
  
  
  $|xy| \leq p$,
  $|y| \geq 1$, and
  for all $i \in \mathbb{N}$, $s' = xy^{i}z \in L$.
  


So if you wanted to show that a language obeyed the pumping lemma (note that is would not prove that it's regular, there are non-regular languages that can also be pumped), you have to first determine the pumping length $p$, then look at every string $s \in L$ and show that there is some way that $s$ can be broken down so that it can be pumped according to the conditions.

So in your case, if we ignore that you don't know $p$, the mistake is showing that one way of breaking it up doesn't work. So assuming that the pumping length is at least $4$ (it is, but why...?), then the breakdown $x = 000$, $y = 1$, $z = 1^{n-1}$ can quite happily be pumped.

When this has settled in, it also shows how the proofs of non-regularity via the pumping lemma work. You just have to find one string, but you have to show that there is no workable breakdown for that string.