Mapping reduction to show NeverHalt is undecidable
You will not prove that $NeverHalt_{TM}$ is undecidable by "finding a TM whose language will "help" us solve" it. This would actually prove the opposite, if $M$ is a TM deciding some language. 

You have to suppose $NeverHalt_{TM}$ to be decidable by a turing machine $D$, and show that with its help you can decide other non-recursive language. In other words, you have to pick a suitable undecidable language and reduce it to $NeverHalt_{TM}$. There is a reduction from the emptiness problem for turing machines : $E_{TM} = \{\left &lt; M \right &gt;\ :\ \mathcal L(M) = \varnothing \}$.