Context-free Language - Pumping Lemma?
The solution which I provide below is very similar to Example 2.37 in classical Sipser's book "Introduction to the theory of computation"


  Let $p,r&gt;0$ be the pumping length given by the lemma.


I think you have a little abuse of notation. Look carefully at definition. It states that if $L$ is $CFL$ then, there exists some constant $p$ (pumping length) that for every string $s \in L$ such that $|s| \ge p$, $s$ can be divided on five parts such that the following properites hold:


  $|vx|≥1$ $|vwx|≤n$ and for every $i≥0, uv^iwx^iy∈L$


So, let us take $s = a^nb^nc^{2n}$, where $n+n+2n &gt; p$. Then the conditions for lemma are hold, and it follows that $s = uvwxy$ with all those properties. From the first property we get that at least one of $v$ or $x$ is non-empty string. Now, we should consider several cases:

1) At least one of $v$ or $x$ contains diffent letters of the alphabet (for example, $v$ = $zabt$, $|z|, |t| \ge 0$). Then $uv^2wx^2y$ doesn't has the form $a^nb^nc^m$ (problems with order, in my example $v^2 = zabtzabt$), so we get a contradiction.

2) Either $v$ or $x$ contains only one letter from alphabet. Here we get such cases:

2.1) $v$ and $x$ don't contain the letter $c$. By lemma the string $s_1 = uv^0wx^0y \in L$. But $s_1$ has less than $n$ letters $a$ or $b$, while it contains $2n$ letters $c$. So, the property $m \le 2n$ doesn't hold in this case and we get a contradiction.

2.2) $v$ or $x$ contain the letter $c$, so either $a$ or $b$ doesn't belong to $v$ and $x$. Then string $s_1 = uv^2wx^2y$ contains at least $2n + 1$ $c$'s, and only $n$ $a$'s or $b$'s. Again, we get a contradiction.