You would say:
$$\begin{eqnarray}
\mathsf{E}\left(W_1(t_1) W_2(t_2)\right) &=& \mathsf{E}\left(\int_0^{t_1} \mathrm{d}W_1(s_1) \int_0^{t_2}\mathrm{d}W_2(s_2)\right) \stackrel{\text{Ito isometry}}{=} \mathsf{E}\left(\int_0^{\min(t_1,t_2)} \rho\, \mathrm{d}t \right) \\ &=& \int_0^{\min(t_1,t_2)} \rho\, \mathrm{d}t = \rho \min\left(t_1, t_2\right)
\end{eqnarray}
$$
If $W_t:=\rho W^{(1)}_t+\sqrt{1-\rho^2} W^{(2)}_t$ then we can show that,
$W_t$ is a Brownian motion.
Proof
Let $(\Omega, \mathcal{F},\mathbb{P},\{\mathcal{F_t}\})$ be a probability space . Clearly, $W_t$ has continuous sample paths and $W_0=0$.
$$\mathbb{E}[W_t|\mathcal{F_s}]=\rho\,\mathbb{E}[W^{(1)}_t|\mathcal{F_s}]+\sqrt{1-\rho^2}\,\mathbb{E}[W^{(2)}_t|\mathcal{F_s}]=\rho W^{(1)}_s+\sqrt{1-\rho^2} W^{(2)}_s=W_s$$
So $W_t$ is a martingale. Now we should show $W_t^2-t$ is a martingale.
By application of Ito's lemma, we have
$$dW_t^2=2W_tdW_t+d[W_t,W_t]$$
$$dW_t^2=2W_tdW_t+\rho^2 d[W_t^{(1)},W_t^{(1)}]+(1-\rho^2) d[W_t^{(2)},W_t^{(2)}]+2\rho\sqrt{1-\rho^2}d[W_t^{(1)},W^{(2)}_t]$$
Since $W^{(1)}_t$ and $W^{(2)}_t$ are two independent Brownian motions, thus $d[W_t^{(1)},W^{(2)}_t]=0$ , hence
$$dW_t^2=2W_tdW_t+dt$$
consequently
$$d(W_t^2-t)=2W_tdW_t+dt-dt=2W_tdW_t$$
so to speak
$$d(W_t^2-t)=2W_tdW_t$$
Indeed
$$d(W_t^2-t)=2\rho W_t dW^{(1)}_t+2\sqrt{1-\rho^2}\,W_t dW^{(2)}_t$$
Therefore $W_t^2-t$, is a martingale (because it's SDE has a null drift ) and $W_t$ is a standard Brownian motion.
Best Answer
I solved and one should use these rules:
.$\left(dt\right)^{n}\in$o$\left(dt\right)$
.$\mathbb{E}[f(B_t)]=f(B_t)$ if $Var[f(B_t)]=0$
.$dB_t^2=dt$
From the result that I already have i.e.
$\mathbb{E}(dBxdBy)=\rho dt$
We just have to show that $Var(dBxdBy)=0$, and apply the second rule, but
$Var(dBxdBy)=\mathbb{E}(dBxdBy)^2-\mathbb{E}^2(dBxdBy)=\mathbb{E}(dtdt)-\rho2dt^2=dt^2-\rho^2dt^2=0$
QED