Proof that the language $ww^R$ is not regular without using the pumping lemma
Let

$$ L_1 = \{\ \omega\omega^r \ | \ \omega \in \Sigma^* \} $$
$$ L_2 = \{\ a^nb^n \ | \ n \in \mathbb{N}\ \}$$

Because $\Sigma = \{a, b, c\}$ is finite, $\Sigma^*$ must be countable.

This means we can define a function $f : \Sigma^* \rightarrow \mathbb{N}$ which returns the position of a word $\omega \in \Sigma^*$ in an arbitrary ordering defined by $f$. $f$ is invertible, we define $f^{-1} : \mathbb{N} \rightarrow \Sigma^*$, which returns the word $\omega$ at position $n$ in the enumeration of $\Sigma^*$ as defined by $f$.

Now, since every word $\omega$ is uniquely defined by a natural number $n$, and every word $\omega$ has exactly one unique reverse $\omega^r$, we can define an isomorphism between $L_1$ and $L_2$:

$$ g \colon\  L_1 \rightarrow L_2; \quad \omega\omega^r \mapsto a^{f(\omega)}b^{f(\omega)} $$

$$ g^{-1} \colon \ L_2 \rightarrow L_1; \quad a^nb^n \mapsto f^{-1}(n) f^{-1}(n)^r $$

Since $L_2$ is not regular and we have a bijection between $L_1$ and $L_2$, $L_1$ must not be regular.

This proof works for any finite alphabet $\Sigma$ because the transitive closure of a finite alphabet is always countable.