I am following the proof of the existence of solutions of SDE: let $b(t,x)$ and $\sigma(t,x)$ be Lipschitz continuous and consider the following SDE
$dX_t=b(t,X_t)dt + \sigma(t,X_t)dB_t$.
Define
$Y_t^{n+1}=X_0 + \int_{0}^{t}b(s,Y^{(n)}_s)ds + \int_{0}^{t}\sigma(s,Y^{(n)}_s)dB_s$.
Then it can be shown that $(Y_t^{(n)})_{n=0}^\infty$ forms a Cauchy sequence and thus
$X_t:= \lim_{n\to\infty}Y_t^{(n)}$ in $L^2(\lambda \times P)$
where $\lambda$ denotes the Lebesgue measure. So from Lipschitz condition we also have that $b(t,Y_t^{(n)})\to b(t,X_t)$ and $\sigma(t,Y_t^{(n)}\to \sigma(t,X_t)$ in $L^2(P)$.
Now the final task is to show that $X_t$ satisfies the original SDE.
So we are left to prove that
$\int_{0}^{t}b(s,Y^{(n)}_s)ds \to \int_{0}^{t}b(s,X_s)ds$ and $\int_{0}^{t}\lambda(s,Y^{(n)}_s)dB_s \to \int_{0}^{t}\lambda(s,X_s)dB_s$ in $L^2(P)$.
However I fail to prove these results. For the first part references suggest that Holder's inequality can be applied, but I I can't see how the result follows from the following
$E\Bigg[\Bigg(\int^{t}_0\big(b(s,Y^{(n)}_s)-b(s,X_s)\big)ds\Bigg)^2\Bigg]$
This does not follow in general from the convergence you have. However, the following convergence does follow: $b(\cdot,Y^{(n)})\to b(\cdot,X)$ and $\sigma(\cdot,Y^{(n)})\to \sigma(\cdot,X)$ in $L^2(\lambda\times P)$.
Then the convergence $$\int_{0}^{t}\sigma(s,Y^{(n)}_s)dB_s \to \int_{0}^{t}\sigma(s,X_s)dB_s$$ in $L^2(P)$ follows from the continuity of Ito integral in $L^2(\lambda\times P)$. The other follows e.g. from the Cauchy-Schwarz inequality (which is of course Holder for $p=q=2$): $$ E\left(\int_0^t \big(b(s,Y_s^{(n)})-b(s,X_s)\big)ds\right)^2 \le t E\left(\int_0^t \big(b(s,Y_s^{(n)})-b(s,X_s)\big)^2ds\right)\\ = t\|b(\cdot,Y^{(n)})- b(\cdot,X)\|^2_{L^2(\lambda\times P)}\to 0,\ n\to\infty. $$