Is the language of Turing Machines that halt on every input recognizable?
Preliminary

Let $H$ be the usual halting language:
$$
H=\{(\langle M\rangle, w)\mid M(w) \text{ halts}\}
$$
Its complement is
$$
\overline{H}=\{(\langle M\rangle, w)\mid M(w) \text{ doesn't halt}\}
$$
We know that $H$ is recognizable (just run $M$ on $w$ and do what it does), so $\overline{H}$ must be unrecognizable (otherwise $H$ would be decidable, which we know isn't true).

The Reduction

As you suggested, we'll produce a reduction from $\overline{H}$ to
$$
AH = \{\langle N\rangle\mid N \text{ halts on all }x\}
$$
thus showing that $AH$ is unrecognizable. As you suggested, what we want is to map $(\langle M\rangle, w)$ to a description of a TM $M_w$ where

Mw(x) =
   run M on w
   if M(W) halts
      loop
   if M(w) doesn't halt
      halt


That would do it, except for the fact that if $M$ didn't halt on $w$, we'd never get to the $\texttt{halt}$ instruction. What we need is a finite way of testing for non-halting. A standard way to deal with this kind of difficulty is to limit the number of steps our simulation will perform, like this:

N(x) =
   run M on w for |x| steps
   if M has halted
      loop
   if M(w) hasn't halted yet
      halt


How does this modification help us?


If $(\langle M\rangle, w)\in\overline{H}$, then $M(w)$ will never halt, so $N(x)$ will halt after $|\,x\,|$ steps and thus $\langle N\rangle\in AH$
If $(\langle M\rangle, w)\notin\overline{H}$ then $M(w)$ will eventually halt, say after $s$ steps. This means that $N$ will halt on all inputs, $x$ of length less than $s$, and will loop on all longer inputs, so $\langle N\rangle\notin AH$.


This establishes the reduction we needed, and so we can conclude that $AH$ is non-recognizable.