Definitions:
$1)$ A process $X$ is adapted to a filtration $(\mathcal{F}_t)$ if $X(t)$ is $\mathcal{F}_t$-measurable for each $t ≥ 0$
$2)$ A process $X$ is $\mathcal{F}_t$-progressive if for each $t ≥ 0$, the mapping $(s,ω) \rightarrow Xs(ω)$ is measurable on $([0,t]×Ω,\mathcal{B}([0,t])⊗\mathcal{F}_t)$.
To proof the main statement: Since $X$ is $\mathcal{F}_t$-progressive then for every $A\in \mathcal{B}$, we have that $$\{(t,\omega):X_t\in A\}\in(\mathcal{B}([0,t])⊗\mathcal{F}_t)$$
and we want to show that for every $A\in \mathcal{B}$, we have $$\{\omega:X_t\in A\}\in\mathcal{F}_t$$
As intuitive as this seems, I'm stuck and can't produce a formal proof - my ideas were mainly related to using the fact the that if $A ∈ \mathcal{F}_1 ⊗ \mathcal{F}_2$, then for each $x_1 ∈ Ω_1$ the section $$A_{x_1} = \{x_2 ∈ Ω_2 : (x_1, x_2) ∈ A\}$$ is measurable but to no success.
Thanks in advance.
Progressive measurability means that the mapping
$$([0,t] \times \Omega, \mathcal{B}[0,t] \otimes \mathcal{F}_t) \ni (s,\omega) \mapsto X_s(\omega) \in (\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$$
is measurable for any $t>0$. On the other hand, it is not difficult to check that the mapping
$$(\Omega,\mathcal{F}_t) \ni \omega \mapsto (t,\omega) \in ([0,t] \times \Omega, \mathcal{B}[0,t] \otimes \mathcal{F}_t)$$
is measurable (recall that $t$ is fixed). This implies that the composition
$$(\Omega,\mathcal{F}_t) \ni \omega \mapsto X_t(\omega) \in (\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$$
is measurable.