How to reduce EQU to UNI?
This answer assumes that $T(M)$ is the set of inputs on which $M$ halts.

Given an instance $u\#v$ of $\texttt{EQU}$, we construct a new Turing machine $w$ with the following input:


A string $x$.
An integer $t$.


The machine acts as follows.


Run $u$ on $x$ for $t$ steps. If $u$ halted, then run $v$ on $x$.
Run $v$ on $x$ for $t$ steps. If $v$ halted, then run $u$ on $x$.
Halt.


I claim that $w \in \texttt{UNI}$ iff $u\#v \in \texttt{EQU}$. To see this, consider some input $x$. We distinguish four different cases:


Both $u$ and $v$ halt on $x$. In this case $w$ will halt on $x,t$ for all $t$ (since "run $v$ on $x$" and "run $u$ on $x$" will always terminate).
Both $u$ and $v$ don't halt on $x$. In this case the test in steps 1–2 will always fail, and so $w$ will always reach step 3 and halt on $x,t$, for any $t$.
$u$ halts on $x$ after $t$ steps, and $v$ doesn't halt on $x$. In this case $w$ won't halt on $x,t$, getting stuck at step 1.
$v$ halts on $x$ after $t$ steps, and $u$ doesn't halt on $x$. In this case $w$ won't halt on $x,t$, getting stuck at step 2.




How did I construct this machine? First of all, let us note that $\texttt{UNI}$ is $\Pi_2$-complete. This means that we can reduce to it any language $L$ such that
$$
x \in L \leftrightarrow \forall y \exists z \, \Pi(x,y,z),
$$
where $\Pi$ is any computable predicate. Next, using $H(u,x,t)$ for "$M_u$ halts on $x$ within $t$ steps", we see that
$$
u\#v \in \texttt{EQU} \leftrightarrow \forall x (\exists s \, H(u,x,s) \land H(v,x,s)) \lor (\forall t \, \lnot H(u,x,t) \land \lnot H(v,x,t)).
$$
Rearranging, this is the same as
$$
u\#v \in \texttt{EQU} \leftrightarrow \forall x \forall t \exists s \, (H(u,x,s) \land H(v,x,s)) \lor (\lnot H(u,x,t) \land \lnot H(v,x,t)).
$$
Therefore $\texttt{EQU}$ should reduce to $\texttt{UNI}$. Taking a look at the $\Pi_2$-completeness proof, we reach the machine outlined above.



It is easy to see that $\texttt{UNI}$ reduces to $\texttt{EQU}$, by mapping $w$ to $w\#h$, where $M_h$ is a machine that always halts. We conclude that $\texttt{UNI}$ is also $\Pi_2$-complete.