Instead of giving a formal argument, I’m going to try to give a clearer picture of what’s really going on and see if that helps you complete the argument.
If $T$ is any tree, and $x\in T$, let $\operatorname{pred}(x,T)=\{z\in T:z, let $\operatorname{ht}(x,T)$ be the height of $x$ in $T$, and for any ordinal $\alpha$ let $\operatorname{Lev}_\alpha(T)=\{x\in T:\operatorname{ht}(x,T)=\alpha\}$.
The whole point of the construction is to replace the Suslin tree $T$ with nicer Suslin tree. The step from $T$ to $T_1$ prunes the ‘dead’ branches of $T$, leaving only nodes that have successors arbitrarily high in the tree.
A ‘nice’ tree branches only at successor levels, so that nodes at limit levels are completely determined by their sets of predecessors. Thus, we’d like $T_1$ to have the property that if $\beta$ is a limit ordinal, $x,y\in\operatorname{Lev}_\beta(T_1)$, and $x\ne y$, then $\operatorname{pred}(x,T_1)\ne\operatorname{pred}(y,T_1)$. Unfortunately, it’s quite possible that $T_1$ does branch at limit levels; the step from $T_1$ to $T_2$ is designed to get rid of this unwanted branching.
Suppose that $\beta$ is a limit ordinal, $x,y\in\operatorname{Lev}_\beta(T_1)$, $x\ne y$, and $\operatorname{pred}(x,T_1)=\operatorname{pred}(y,T_1)$. Let $C=\operatorname{pred}(x,T_1)$; then $C$ is a chain of limit length in $T_1$, so in $T_2$ we have a vertex $a_C$. Where does it fit into $T_1$? Clearly it lies above every $z\in C$, but it lies below both $x$ and $y$. If it were the only new vertex, it would clearly be in $\operatorname{Lev}_\beta(T_2)$, immediately below $x$ and $y$, which would be pushed up from $\operatorname{Lev}_\beta(T_1)$ to $\operatorname{Lev}_{\beta+1}(T_2)$. All successors of $x$ and $y$ at heights $\beta+n$ for $n\in\omega$ would also be pushed up one level, but there would be no change at heights $\ge\beta+\omega$.
Of course we’ve no reason to suppose that this $a_C$ is the only new vertex. However, this turns out not to be a problem if we look at the construction in steps rather than all at once. Imagine climbing $T_1$ and examining each limit level in turn. Say that you’ve reached $\operatorname{Lev}_\beta(T_1)$ for some limit ordinal $\beta<\omega_1$. You find all of the unwanted splits at Level $\beta$ (if any) and treat them as we did the single split in the last paragraph. This does not change the structure of the tree below Level $\beta$ in any way, so it has no effect on changes that we’ve already made lower in the tree. It does ensure that distinct vertices in $\operatorname{Lev}_\beta(T_2)$ have distinct sets of predecessors in $T_2$. And it has no effect on the structure of the tree at heights $\ge\beta+\omega$. In particular, it has no effect on the higher limit levels. Thus, we can repair the unwanted splits one limit level at a time, knowing that repairs at one level can neither affect earlier repairs at lower levels nor affect the unwanted splits at higher levels.
Now the construction that you specified in the problem is just a little different from the one that I just described. I added new elements at limit levels only where that was necessary in order to repair an unwanted split. Your construction is actually going to replace the whole limit level with new elements, whether they’re actually needed or not. This is less intuitive, but it makes the process a little easier to handle, because it makes it more consistent. Your construction amounts to doing the following at a limit level $\beta$.
For $x,y\in\operatorname{Lev}_\beta(T_1)$ write $x\sim y$ iff $\operatorname{pred}(x,T_1)=\operatorname{pred}(y,T_1)$; clearly $\sim$ is an equivalence relation on $\operatorname{Lev}_\beta(T_1)$. For each $\sim$-class $[x]$ let $C([x])=\operatorname{pred}(x,T_1)$, a chain of limit length in $T_1$. Let $\operatorname{Lev}_\beta(T_2)=\{a_{C([x])}:x\in\operatorname{Lev}_\beta(T_1)\}$. Then fix up the order as you described in your construction.
When a vertex $x$ is the only member of its $\sim$-class, it wasn’t part of an unwanted split, so the new vertex $a_{C([x])}$ isn’t actually necessary. The nice thing, though, is that this uniform construction simply pushes the next $\omega$ levels of $T_1$ up a notch: for each $n\in\omega$ we have $\operatorname{Lev}_{\beta+n+1}(T_2)=\operatorname{Lev}_{\beta+n}(T_1)$. All we’re doing is replacing the limit levels in a way that kills off any splits at levels that may have been present in $T_1$.
The final step, from $T_2$ to $T_3$, is intended to get rid of ‘fake’ nodes, i.e., nodes at which no branching occurs. (After all, if there’s no branching there, why have a vertex there?) This can indeed affect the heights of vertices and hence the composition of the levels; a vertex on Level $\omega$ of $T_2$ could end up on Level $1$ of $T_3$, for instance. Note, though, that all level changes are downward: $\operatorname{ht}(x,T_3)\le\operatorname{ht}(x,T_2)$ for all $x\in T_3$.
(I need to run off and take care of a few chores. I’ll post this much for now, since I think that it’s a useful start for you, and finish it off in a bit.)