In a proof of a Theorem I have seen the following reasoning. Let $\sigma_s$ be a cadlag process. Consider the integral
$$ I(\Delta) = \int_0^\Delta\sigma_s\,dW_s $$
where $W$ is Brownian motion independent of $\sigma$. Then, it is said that by stochastic Taylor expansion it holds that
$$ I(\Delta) = \sigma_0\,(W_{\Delta}-W_0)+O_p(\Delta^{1/2}) \tag{1}. $$ It is not clear to me how a stochastic Taylor expansion leads to equation (1).
In order to show that
$$I(t) := \int_0^t \sigma_s \, dW_s$$
satisfies
$$I(t) = \sigma_0 (W_t-W_0) + O_p(\sqrt{t}) \quad \text{as $t \to 0$} \tag{1}$$
we have to prove that
$$\frac{1}{\sqrt{t}} \bigg[ I(t) - \sigma_0 (W_t-W_0) \bigg]$$
is bounded in probability, i.e.
$$\forall \epsilon>0 \, \, \exists \delta>0, M>0 \, \, \forall t \in [0,\delta]: \quad \mathbb{P} \left( \left| \frac{1}{\sqrt{t}} \bigg[ I(t) - \sigma_0 (W_t-W_0) \bigg]\right| > M \right) \leq \epsilon. \tag{2}$$
Clearly,
$$I(t)-\sigma_0(W_t-W_0) = \int_0^t (\sigma_s-\sigma_0) \,dW_s.$$
Applying the Markov inequality and using Itô's isometry, we find
$$\begin{align*} \mathbb{P} \left( \left| \frac{1}{\sqrt{t}} \bigg[ I(t) - \sigma_0 (W_t-W_0) \bigg]\right| > M \right) &\leq \frac{1}{M^2 t} \mathbb{E} \left( \left| \int_0^t (\sigma_s-\sigma_0) \, dW_s \right|^2 \right) \\ &= \frac{1}{M^2 t} \mathbb{E} \left( \int_0^t (\sigma_s-\sigma_0)^2 \,ds \right). \end{align*}$$
Since $\sigma$ is, by assumption, a bounded process, this implies that there exists a finite constant $K>0$ such that
$$\mathbb{P} \left( \left| \frac{1}{\sqrt{t}} \bigg[ I(t) - \sigma_0 (W_t-W_0) \bigg]\right| > M \right) \leq \frac{K}{M^2}$$ for all $t \in [0,1]$. Choosing $M>0$ sufficiently large we get $(2)$.