How to show $\frac{1}{\delta} \mathbb{P}[ \sup_{t\leq s \leq t+\delta} |X(s)-X(t)|\geq \epsilon] \leq \eta$? If a random element $X$ of $D[0,1]$ has the property that
$$\lim_{\delta\to 0} \sup_{0\leq t\leq 1-\delta} \frac{1}{\delta}\mathbb{P}(|X(t+\delta)-X(t)|\geq \epsilon) = 0.$$
for every $\epsilon >0$, then $\mathbb{P}(X\in C[0,1])=1$.
$D[0,1]$ be the space of real functions $f$ on $[0,1]$ that are right continuous and have left limit. That is space of cadlag functions. $C[0,1]$ is set of continuous function.

My solution:
$$\lim_{\delta\to 0} \sup_{0\leq t\leq 1-\delta} \frac{1}{\delta}\mathbb{P}(|X(t+\delta)-X(t)|\geq \epsilon) = 0.$$
implies $\forall \ \epsilon >0$ and $\forall \ \eta>0$, $\exists \delta\in(0,1)$ such that
$$\sup_{0\leq t\leq 1-\delta} \frac{1}{\delta}\mathbb{P}(|X(t+\delta)-X(t)|\geq \epsilon) \leq \eta.......(*)$$
I have to show that
1.For each $\eta >0$ $\exists a$ such that $\mathbb{P}(|X(0)|>a)\leq \eta$
2.For each $\epsilon >0$ and $\eta >0$ $\exists \delta \in (0,1)$ such that
$\mathbb{P}(w(X,\delta) \geq \epsilon) \leq \eta$. where $w(x,\delta) = \sup_{0\leq t \leq 1-\delta} |x(t+\delta) - x(t)|$.
First claim is trivial since $X\in D[0,1]$ hence $||x||_{\infty} < \infty$ hence the first claim.
For the second claim, I want to used the theorem 7.4 from the Billingsley's book Convergence of probability measures (2nd edition).

Theorem 7.4: Suppose that $0=t_0 <t_1<....<t_v=1$ and $$\min_{i=1,...,v}(t_i - t_{i-1}) \geq \delta.$$ Then, for arbitrary x,
$$w(x,\delta) \leq 3\max_{i=1,...,v}\sup_{t_{i-1}\leq s \leq t_i} |x(s)-x(t_i)|$$
and, for arbitrary $\mathbb{P}$,
$$\mathbb{P}(x| w(x,\delta)\geq 3\epsilon) \leq \sum_{i=1}^v \mathbb{P}[x| \sup_{t_{i-1}\leq s \leq t_i} |x(s)-x(t_i)|\geq \epsilon]$$

Using this theorem and $$\frac{1}{\delta} \mathbb{P}[x| \sup_{t\leq s \leq t+\delta} |x(s)-x(t)|\geq \epsilon] \leq \eta$$. We can prove second claim.
If I prove the second claim then we are able to show our desired proof using the theorem stated in the same book 1st edition (stated below).

Theorem: Let $\{\mathbb{P}_n\}$ be defined on $(D[0,1], \sigma(D[0,1]))$. Then $\{\mathbb{P}_n\}$ is tight if
1.For each $\eta >0$ $\exists a$ such that $\mathbb{P}_n(|X(0)|>a)\leq \eta \ \forall n$ and,
2.For each $\epsilon >0$ and $\eta >0$ $\exists \delta \in (0,1)$ and $n_0$ such that
$\mathbb{P}_n(w(X,\delta) \geq \epsilon) \leq \eta \ \forall n>n_0$. If a sub-sequence $\{\mathbb{P}_{n'}\}$ converges weakly to $\mathbb{P}$, then $\mathbb{P}(C[0,1])=1$.

But I don't know how to show the $$\frac{1}{\delta} \mathbb{P}[ \sup_{t\leq s \leq t+\delta} |X(s)-X(t)|\geq \epsilon] \leq \eta$$ form $(*)$.
Other ingenious approaches are also appreciated.
 A: I will sketch a proof of a more general result.

Proposition If a càdlàg$^*$ process $X$ is such that
$$
\sum_{k=1}^n \mathbb{P}\big(|X({k/n}) - X((k-1)/n)|> \epsilon\big)\to 0, n\to\infty,\tag{1}
$$
for any $\epsilon>0$, then $X\in C[0,1]$ almost surely.

Note that your assumption clearly implies (1), since
$$
\sum_{k=1}^n \mathbb{P}\big(|X({k/n}) - X((k-1)/n)|> \epsilon\big) \le n \sup_{0\le t\le 1-1/n} \mathbb{P}\big(|X(t+1/n) - X(t)|> \epsilon\big).
$$
Proof of Proposition Let $N_{\epsilon}$ be the number of jumps of $X$ greater than $\epsilon$ and $N_{\epsilon,n}$ be the number of $k\le n$ such that $|X(k/n) - X((k-1)/n)|\ge \epsilon/2$. Since $X$ is càdlàg, $N_\epsilon\le \liminf_{n\to\infty} N_{\epsilon,n}$.
By the Fatou lemma,
$$
\mathbb E\Big[\liminf_{n\to\infty} N_{\epsilon,n}\Big]\le \liminf_{n\to\infty}\mathbb E\Big[ N_{\epsilon,n}\Big]  = \liminf_{n\to\infty} \mathbb E\bigg[ \sum_{k=1}^n \mathbb{1}_{|X(k/n) - X((k-1)/n)|\ge \epsilon/2}\bigg]\\
= \liminf_{n\to\infty} \sum_{k=1}^n\mathbb P\big( |X(k/n) - X((k-1)/n)|\ge \epsilon/2\big) = 0. 
$$
Therefore, $\liminf_{n\to\infty} N_{\epsilon,n} = 0$ almost surely and hence $N_\epsilon = 0$ almost surely for any $\epsilon >0$. It follows that $X$ is continuous almost surely.

$^*$ It suffices to assume that $X$ is separable.
