I wrestled with this question for an annoyingly long time. Now understanding things much better, I'm a bit embarrassed that this took me so long to figure out (such is the process of learning though). I was quite frustrated that I couldn't find the details of this anywhere, aside from some helpful comments in this paper of Ivić. To spare others my frustration, I'm posting the answer in a Q&A style, as well as for my own reference.

It suffices to show that if $V > 0$ and $t_1,\ldots, t_L \in [T,2T]$ with $t_{l+1}-t_l \geq 2$ and $|\zeta(\frac{1}{2}+it_l)| \geq V$, then
$$
L \ll T^{2+\varepsilon}V^{-12},
$$
for then the estimate for the twelfth moment follows by the argument originally given by Heath-Brown in his paper first proving the estimate (2). Note that we may also assume that $V > T^{1/8+\varepsilon/2}$, since the contribution to the integral (2) of the $t$ with $|\zeta(\frac{1}{2}+it)| \leq T^{1/8+\varepsilon/2}$ is
$$
\leq T^{1+4\varepsilon} \int_{T}^{2T} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{4} dt \ll T^{2+5\varepsilon}
$$
by the standard estimate for the fourth moment of $\zeta$. In particular, we may assume $V^4 > T^{1/2+2\varepsilon}$.

Cover the interval $[T,2T]$ by $R$ disjoint intervals $I_1,\ldots,I_{R}$ of length $V^4T^{-2\varepsilon}$ ($> T^{1/2}$) such that each point $t_l$ lies in some interval $I_r$. Then
$$
R \leq L \leq \sum_{r=1}^R \sum_{t_l \in I_r} \frac{\left|\zeta\left(\tfrac{1}{2}+it_l\right) \right|^4}{V^4}.
$$
At this point, we need that for $t \in [T,2T]$,
$$
\tag{3}
\left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{2k} \ll \log T \int_{t-1/3}^{t+1/3} \left|\zeta\left(\tfrac{1}{2}+iu\right) \right|^{2k} du,
$$
which can be found in the paper of Ivić linked above. Since the points $t_l$ are spaced by at least 2,
$$
\sum_{t_l \in I_r} \left|\zeta\left(\tfrac{1}{2}+it_l\right) \right|^4 \ll \log T \int_{I_r'} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt,
$$
where $I_r'$ is the interval $I_r$ enlarged by $1/3$ on either end.
Applying the estimate (1), we find that
\begin{align*}
\sum_{r=1}^R \log T \int_{I_r'} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt &\ll \left(RV^4 T^{-2\varepsilon} + R^{1/2}V^{-2}T^{1+\varepsilon} \right)T^{\varepsilon} \\
&\ll R T^{-\varepsilon} V^4 + R^{1/2}T^{1+2\varepsilon}V^{-2}.
\end{align*}
and so in particular
$$
\tag{4}
R \leq L \ll RT^{-\varepsilon} + R^{1/2}T^{1+2\varepsilon}V^{-6}
$$
Since $RT^{-\varepsilon} < cR$ for any constant $c > 0$ and $T$ sufficiently large, we must have
$$
R\ll R^{1/2}T^{1+2\varepsilon}V^{-6}.
$$
Thus
$$
R \ll T^{2+4\varepsilon} V^{-12},
$$
and the required bound for $L$ now follows from (4).