= Solution
For each sample point put $A_n=\max_{0\leq k<2^n}|X_{(k+1)2^{-n}}-X_{k2^{-n}}|$. First take $s<t$ in the <dyadic rationals>, set $h=t-s$, and choose $N\geq0$ with $2^{-N}\leq h<2^{1-N}$. The left grid approximations $s_N=2^{-N}\lfloor2^Ns\rfloor$ and $t_N=2^{-N}\lfloor2^Nt\rfloor$ are at most two grid steps apart, so $|X_{t_N}-X_{s_N}|\leq2A_N$.
For either endpoint $r$, successive left approximations $r_n,r_{n+1}$ are equal or one step apart on the level-$(n+1)$ grid. Since $r$ is dyadic these approximations eventually equal $r$ exactly. The <triangle inequality> therefore gives
$$
|X_t-X_s|\leq2\sum_{n\geq N}A_n
\leq2^{-N\alpha}\,2\sum_{n\geq N}2^{n\alpha}A_n
\leq h^\alpha K_\alpha.
$$
The second inequality is valid also for $\alpha=0$. Thus \b[the required bound holds simultaneously for every dyadic pair] whenever $K_\alpha$ is finite; if it is infinite the inequality holds in the extended sense. For $s=t$ the left side is zero. This is <dyadic increment chaining>, with the stated constant two.
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