Decidability of the language that accepts a universal turing machine
If $L_{u}$ would be decidable,  $L_u^\complement$ would be decidable too (the complement of a recursive language is recursive too) , but if you can build a TM that accepts $L_u^\complement$ you can build a TM that accepts $L_d=\{w \mid w_i \notin L(M_i)\} $ that is not RE.

In fact if $w \in L_d  \implies (w,w)\notin L_u \implies (w,w) \in L_u^\complement$.

So with a reduction from $L_d$ to $L_u$ it can be shown that $L_u$ is not decidable.