I looked through many sources which give this as an example for cfl. It also makes sense according to this:
But it fails the pumping lemma test. Let's take n=5. According to the Pumping Lemma, we can decompose s into uvxyz like: u=0, v=0, x=0,y=01, z=111
If we pump it, it clearly fails. So, is 0^n1^n context free or not?
You have just shown that the given language does not meet the condition of the pumping lemma for $n=5$ and one particular decomposition.
To conclude that $L$ is not context free you need to show that the language does not meet the condition of the pumping lemma for some word of sufficiently large length $n$ and all decompositions of such word (within the constraints of the pumping lemma).
It turns out that the words of $L = \{0^n 1^n \mid n \ge 0\}$ admit a valid decomposition as soon as $n \ge 1$. Here is a decomposition of $0^5 1^5$ into $uvxyz$ that can be pumped: $u=0^4$, $v=0$, $x=\varepsilon$, $y=1$, $z=1^4$.
This is not surprising, since $L$ is indeed context-free as the argument in your question shows.
