Consider a sequence of predictable processes $H_n$ converging to a predictable process $H$, each of which is bounded.
Consider $$E \left[ \sup_{t \in [0,T]} \left| \int_0^t (H_n(s)-H(s)) dM(s) \right|^2 \right] \leq 4 E \left[ \left| \int_0^T (H_n(s)-H(s)) dM(s) \right|^2 \right] $$ where M is an $L^2$ martingale.
Where does the factor $4$ comes from? Wouldn't Doob's maximal inequality would only yield a factor of $2$?
The factor of $4$ comes from Doob's maximal inequality. For a martingale $X_t$, the inequality is usually written in the following form: $$|| \sup_{t \in [0,T]} X_t ||_p \leq \frac{p}{p-1} || X_T ||_p \qquad (\star )$$
In your case, $p = 2$, so that applying $(\star )$ for the martingale $X_t = \int_0^t H_n(s) - H (s) dM(s)$ we get (in expectation notation): $$\sqrt{ E \left[ \sup_{t \in [0,T]} \left| \int_0^t H_n(s) - H (s) dM(s) \right|^2 \right] } \leq \frac{2}{2-1}\sqrt{ E \left[ \left| \int_0^T H_n(s) - H (s) dM(s) \right|^2 \right] }$$
Squaring both sides you get the factor of $4$:
$$E \left[ \sup_{t \in [0,T]} \left| \int_0^t H_n(s) - H (s) dM(s) \right|^2 \right] \leq 4E \left[ \left| \int_0^T H_n(s) - H (s) dM(s) \right|^2 \right] $$