Correct application of the CFL Pumping Lemma
The pumping lemma for CFL states that if $L$ is a CFL, there exists a constant $N \ge 1$ such that for all $\omega \in L$ with $\lvert \omega \rvert \ge N$ there is a division $\sigma = u v w x y$ with $\lvert v w x \rvert \le N$ and $v x \ne \varepsilon$ such that $u v^k w x^k y \in L$ for all $k \ge 0$.
(The pumping lemma for regular languages is similar, the discussion below applies with minor changes).
You want to prove by contradiction that $L$ is not a CFL. So you assume $L$ is a CFL, and get a contradiction to the lemma. This means, in detail:

As $L$ is a CFL, it satisfies the lemma. In particular, there is a constant $N$ such that...
As the lemma states all long strings in $L$ can be divided, you are free to pick one that is easy to work with.
The lemma states there is some division of $\omega$ such that.... for all $k \ge 0$...; to contradict the lemma you have to prove no division works. In practice, you take a given division of your $\omega$ (any division) and for it pick some $k$ for it that doesn't give a string in the language. You might need to divide this into cases.

What happens for strings that don't belong to the language, or are shorter than $N$, is completely irrelevant. It might be that some strings can be pumped, that some values of $k$ work for all strings, ... What is crucial is that for one string $\omega$ (the one you picked) no division works with one value of $k$ (again, the one you picked). As the lemma asserts it works for all long enough strings, with some division for all $k$, one counterexample is enough. Infinite examples prove nothing.