Given a Brownian motion $\{W_t\}_{t\in[0;T]}$ and a continuous, adapted and square-integrable (bounded if you want) process $\{\sigma_t\}_{t\in[0;T]}$ and $\varepsilon > 0$, I want to prove that there is a $\delta > 0$ such that:
For all $s \in [0;T]$ and all $M \in \mathcal F_s$, it is
$$ \mathbb E\bigg( 1_M \max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg| \bigg) \le \varepsilon \cdot \mathbb P(M). $$
For $\sigma \equiv 1$, this is easy because we just consider $$ \mathbb E\Big(1_M \max_{s \le t \le (s+\delta) \wedge T} |W_t - W_s|\Big) $$ for which we have a bound due to the distribution of the running maximum of $W$ and the increments of the Brownian motion are independent from the past.
Is there a similar thing for arbitrary Ito integrals (or those satisfying some assumptions)?
Edit: The question has been edited to ask for something stronger which this answer doesn't give. I'm leaving this here in case it inspires someone else/I work out a way to improve it to give the stronger claim.
Since $0 \leq 1_M \leq 1$, we really want to show that we can pick $\delta > 0$ to guarantee that for all $s \in [0,T]$ $$\mathbb{E}\bigg[\max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg| \bigg] \le \varepsilon.$$
We can do this using an appropriate form of Doob's Martingale inequality. We have \begin{align} \mathbb{E}\bigg[\max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg| \bigg] \leq& \mathbb{E}\bigg[\max_{s \le t \le (s+\delta) \wedge T} \bigg| \int_s^t \sigma_u \mathrm dW_u \bigg|^2 \bigg]^{\frac12} \\ \leq& 2\mathbb{E} \bigg[ \bigg(\int_s^{(s+ \delta) \wedge T} \sigma_u dW_u \bigg)^2 \bigg]^{\frac12} \\ =& 2\mathbb{E} \bigg[ \int_s^{(s+ \delta) \wedge T} \sigma_u^2 du \bigg]^{\frac12} \end{align} where the first inequality is just monotonicity of $L^p$-norms on probability spaces, the second is Doob's inequality and the last line then follows by the Ito isometry. It's clear that your assumptions on $\sigma$ let you choose $\delta$ to make this last quantity as small as you like, giving the desired result.