$ \{X_{t}\}_{t\geq 0}, \{Y_{t}\}_{t\geq 0} $ - Ito processes.
From Ito's formula we have got: $$ X_{t}Y_{t} = X_{0}Y_{0} + \int_{0}^{t}X_{s}dY_{s} + \int_{0}^{t}Y_{s}dX_{s} + \int_{0}^{t}dX_{s}dY_{s} \space \space \space \space (1)$$
Using this equality prove that the process $$Z_{t} = (\int_{0}^{t}f(s)dW_{s}\int_{0}^{t}g(s)dW_{s}) - \int_{0}^{t}f(s)g(s)ds $$ is a local martingale for each $f, g \in \mathcal{P}^{2}_{[0, T]}$
Let $X_{t}$ be $\int_{0}^{t}f(s)dW_{s}$ and $Y_{t}$ be $\int_{0}^{t}g(s)dW_{s}$.
$Z_{t} = X_{t}Y_{t} - \int_{0}^{t}g(s)dW_{s}$
Lebesgue integral is constant with respect to omega, so it cannot broke being (local) martingale property for $Z_{t}$. Sum of local martingales is a martingale. So if I show that each integral from $(1)$ is a local martingale I will get that $Z_{t}$ is also. But how to do it? Or maybe there is another better way?
The correct formula is: \begin{equation} X_{t}Y_{t} = X_{0}Y_{0} + \int_{0}^{t}X_{s}dY_{s} + \int_{0}^{t}Y_{s}dX_{s} + \int_{0}^{t}d[X,Y]_s, \end{equation} where $[.,.]_t$ is the quadratic covariation process.
Indeed setting $X_t = \int_{0}^{t} f(s)dW_s$ and $Y_t = \int_{0}^{t} g(s)dW_s$ is a good idea. Then, these integrals satisfy $X_0=Y_0 =0$, whereas $dX_t = f(t)dW_t$, $dY_t = g(t)dW_t$. By the inherent characterisation of stochastic integration,
\begin{equation} [X,Y]_t = \left[\int_{0}^{\cdot} f(s)dW_s, \int_{0}^{\cdot} g(s)dW_s\right]_t = [f \bullet W, g \bullet W]_t = (fg \cdot [W,W])_t = \\(fg \cdot t) = \int_{0}^{t} f(s)g(s)ds, \end{equation} where $\bullet$ denotes integration with respect to a semimartingale and $\cdot$ is integrating with respect to a finite variation process. Therefore: \begin{equation} Z_t = \int_{0}^{t}X_{s}dY_{s} + \int_{0}^{t}Y_{s}dX_{s} = \int_{0}^{t}\left( g(s)X_{s} + f(s)Y_s \right)dW_{s}, \end{equation} which is a continuous local martingale.