Pumping Lemma Proof (Type of wcw language)
The structure of the language i think can be explained as (based on definition above) $\{a^n b^n ~c~ a^n b^n |c \in \{a,b\} \}$

As explained in the comments, this is a wrong inference. If
$$
L = \left \{ d k d ~|~ d \in \{ a, b \}^{*}, k \in \{a, b\} \right \}
$$
then $d$ can be any finite sequence (of any length) of $a$ or $b$ terminals. Recall that, for an alphabet $\Sigma$, Kleene star is defined inductively as
\begin{align}
\Sigma^0  &amp; = \{ \epsilon \} \\
\Sigma^n  &amp; = \{ \omega a ~|~ \omega \in \Sigma^{n - 1}, a \in \Sigma \} \\
\Sigma^{*}  &amp; = \bigcup_{n \ge 0}{\Sigma^n}
\end{align}
So for example
$$
\{ \epsilon, a, b, ab, ba, abab, abbb, baba \} \subset \{ a, b \} ^{*}
$$
while instead
$$
\{ a^n b^n | n \ge 0 \} = \{ \epsilon, ab, aabb, aaabbb, \ldots \}
$$
To tackle your problem, I can recommend negating the Pumping lemma and then

Fix an arbitrary $N &gt; 0$
Take $\omega = a^N b a^N$ with $d = a^N$ and $k = b$
Equate $\omega = xyz$ and note that it must be $|xy| \le N$ and $|y| &gt; 0$

Can you conclude your proof?

 We just chose our $\omega \in L$. We must show that, whatever our decomposition, $\exists k \ge 0. ~ x y^k z \not \in L$. Let $x, y, z$ be such that \begin{cases} \omega &amp; = xyz \\ |xy| &amp; \le N \\ |y| &amp; &gt; 0 \end{cases} then, whatever $y$, for $k = 0$ you always have \begin{equation} x y^0 z = a^M b a^N \end{equation} with $M \neq N$.  Thanks @plop for strenghening the proof.