How to show that process $\int_0^t(g(s)X_1(s)+f(s)X_2(s))dWs$ is a local martingale where
$X_1(s)=\int_0^sf(u)dW_u$ and $X_2(s)=\int_0^sg(u)dW_u$?
where $f,g\in P_{[0,T]}^2=\{f:[0,T]\times \Omega\to \mathbb{R}, f-adapted, \int_0^Tf^2(s)ds<\infty a.s\}$
I would go like this, let's see if $\int_0^T(g(s)X_1(s)+f(s)X_2(s))^2ds<\infty ~ a.s$
So this would be true if for example $A= \mathbb E[\int_0^T(g(s)X_1(s)+f(s)X_2(s))^2ds]$ is finite, and because of classical theorem for local martingale property of stochastic integral for well behaved integrand (check George Lowther's website "almost sure"). So let's look at this :
$$A = \mathbb E[\int_0^T(g(s)X_1(s))^2ds]+ \mathbb E[\int_0^T2 g(s)X_1(s).f(s)X_2(s)ds]+ \mathbb E[\int_0^T(f(s)X_2(s))^2ds]= A_1+ A_{12} +A_2$$ with obvious notations. Now $$A_1= \int_0^T g(s)^2.\mathbb E[X_1(s)^2]ds=\int_0^T g(s)^2.\mathbb E[(\int_0^s f(u)dB_u)^2]ds=\int_0^T g(s)^2.(\int_0^s f(u)^2du)ds\leq \int_0^T g(s)^2.(\int_0^T f(u)^2du)ds=(\int_0^T f(s)^2 ds).(\int_0^T g(s)^2ds)$$ Where we have used Itô's isometry and the fact that $\int_0^s f(u)du$ for positive $f$ is an increasing function in $s$.
The inequality for $A_1$ also holds by symmetry of the argument by permuting $f$ and $g$ for $A_2$. Last for $$A_{12}= 2\int_0^T g(s).f(s).\mathbb E[X_1(s).X_2(s)]ds=2\int_0^T( g(s).f(s).\mathbb E[(\int_0^s g(u)dB_u).(\int_0^s f(u)dB_u)])ds=2\int_0^T g(s).f(s).(\int_0^s g(u)f(u)du) ds\leq 2\int_0^T g(s).f(s).(\int_0^s g(u)^2 du)^{1/2})(\int_0^s f(u)^2du)^{1/2}) ds\leq 2\int_0^T g(s).f(s)ds.(\int_0^T g(u)^2 du)^{1/2})(\int_0^T f(u)^2 du)^{1/2})=2\int_0^T g(s).f(s)ds.(\int_0^T g(u)^2 du)^{1/2}.(\int_0^T f(u)^2 du)^{1/2}\leq 2 (\int_0^T g(s)^2 ds)^{1/2}(\int_0^T f(s)^2 ds)^{1/2}.(\int_0^T g(u)^2 du)^{1/2}.(\int_0^T f(u)^2 du)^{1/2}=2 (\int_0^T g(s)^2 ds)(\int_0^T f(s)^2 ds))$$ Here we used the same properties as above together with 2 applications of the Cauchy-Schwarz inequality.
All this stuff is finite so $A$ is and the process $\int_0^T(g(s)X_1(s)+f(s)X_2(s))dB_s$ is a local martingale.