= Solution
We give a finite <chaining> argument, keeping the constants explicit. If $\delta=0$, the <metric space> has just the point $t_0$ and both sides vanish. Suppose $\delta>0$, and write
$$
\varepsilon_k=2^{-k}\delta,\qquad N_k=N(\varepsilon_k;T,d),\quad k\geq1.
$$
Choose a smallest <metric net> $T_k\subseteq T$ of radius $\varepsilon_k$, and maps $\pi_k:T\to T_k$ satisfying $d(t,\pi_k(t))\leq\varepsilon_k$. Set $T_0=\{t_0\}$ and $\pi_0(t)=t_0$. Since $T$ is finite, choose $K$ with $\varepsilon_K$ smaller than the minimum distance between distinct points; then $T_K=T$ and $\pi_K(t)=t$.
For each $k$, retain only the pairs actually occurring along these approximation paths:
$$
E_k=\{(\pi_k(t),\pi_{k-1}(t)):t\in T\},\qquad m_k=|E_k|.
$$
The <triangle inequality> gives $d(u,v)\leq\varepsilon_k+\varepsilon_{k-1}=3\varepsilon_k$ for $(u,v)\in E_k$; for $k=1$ the same argument uses $d(t,t_0)\leq\delta=\varepsilon_0$. Also
$$
m_1\leq N_1,\qquad m_k\leq N_kN_{k-1}\leq N_k^2\ (k\geq2),\qquad \log m_k\leq2\log N_k.
$$
Here only monotonicity in the radius for the same space $T$ is used. No monotonicity under subsets is assumed.
For every pair in $E_k$, the increment has the <Sub-Gamma random variable in the right tail> bound with common parameters $9\nu\varepsilon_k^2$ and $3c\varepsilon_k$: for $0<s<(3c\varepsilon_k)^{-1}$,
$$
\log\mathbb E e^{s(X_u-X_v)}
\leq\frac{\nu d(u,v)^2s^2}{2(1-cd(u,v)s)}
\leq\frac{9\nu\varepsilon_k^2s^2}{2(1-3c\varepsilon_ks)}.
$$
Identical pairs give the zero increment, which obeys the same bound. Both orientations have finite <exponential moments>, so all increments are <integrable random variables>. Differentiating their two upper bounds at zero also gives $\mathbb E(X_u-X_v)=0$. Applying the maximum bound from part (b), including its $m_k=1$ case, gives
$$
\begin{aligned}
\mathbb E\max_{(u,v)\in E_k}(X_u-X_v)
&\leq3\varepsilon_k\sqrt{2\nu\log m_k}+3c\varepsilon_k\log m_k\\
&\leq6\sqrt\nu\,\varepsilon_k\sqrt{\log N_k}+6c\varepsilon_k\log N_k.
\end{aligned}
$$
The <telescoping sum> along each path is exact:
$$
X_t-X_{t_0}=\sum_{k=1}^K(X_{\pi_k(t)}-X_{\pi_{k-1}(t)}).
$$
Take the <supremum>, bound it by the sum of the maxima, and take <expected values>. This proves
$$
\mathbb E\sup_{t\in T}(X_t-X_{t_0})
\leq6\sqrt\nu\sum_{k=1}^K\varepsilon_k\sqrt{\log N_k}
+6c\sum_{k=1}^K\varepsilon_k\log N_k.
$$
For either $f(u)=\sqrt{\log N(u;T,d)}$ or $f(u)=\log N(u;T,d)$, the function is nonnegative and nonincreasing. Thus
$$
\int_{\varepsilon_{k+1}}^{\varepsilon_k}f(u)\,du
\geq\frac{\varepsilon_k}{2}f(\varepsilon_k),\qquad
\sum_{k=1}^K\varepsilon_k f(\varepsilon_k)
\leq2\int_0^{\delta/2}f(u)\,du.
$$
Combining the estimates yields the required <metric entropy> bound:
$$
\boxed{\mathbb E\left[\sup_{t\in T}X_t-X_{t_0}\right]
\leq12\sqrt\nu\int_0^{\delta/2}\sqrt{\log N(u;T,d)}\,du
+12c\int_0^{\delta/2}\log N(u;T,d)\,du.}
$$
In particular, \b[$C=12$ and $C'=12$ work]. The square-root and linear <metric entropy> terms arise respectively from the two terms in the sub-Gamma maximum bound. Because the chain ends exactly at $T$, no limiting sample-path regularity assumption is needed.
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