Proving non-regularity of $u u^R v$?
When you are stuck in a place like this, it pays to check if the method can work at all. That is, try to prove the opposite. Since the Pumping lemma does not characterise REG, this is actually possible.

Claim: Your language $L$ fulfills the conditions of the Pumping lemma with $p=1$.

Let $w = uu^Rv$ for $u, v \in \Sigma^+$; say $|w|=n$ and $|u|=m$. With the (canonical) notation from the Pumping lemma, pick $x = \varepsilon$, $y=w_1$ and $z = w_2\cdots w_n$.

By definition, $|y| \geq 1$ and $|xy| \leq p$. Furthermore, 


$xy^0z = (u_2\cdots u_m) \cdot (u_2\cdots u_m)^R \cdot u_1 v \in L$,     
$xyz = w \in L$, and
$xy^iz = u_1 u_1^R (u_2 \cdots u_m) u^R v \in L$, for $i \geq 2$.


Q.e.d. -- $L$ has the Pumping property.

That is, the Pumping lemma can not be used here, you need other methods.