$\prod A_\lambda$ is closed proof 
Given a topological product $\prod X_{\lambda}$ and $A_{\lambda}\subset X_{\lambda}$ subspaces of $X_{\lambda}$. Prove the product of closed sets $A_{\lambda}$ is closed.

I tried to prove this in the following way.
Consider an arbitrary $x\in\prod X_{\lambda}$ and $x_{u}\in X_{u}$ in which $u\in\lambda$. I am going to prove the complement is open $x_{\lambda}\notin A_{\lambda},\:\forall \lambda$. As $A_{\lambda}$ is assumed closed, then if $U_\lambda$ is the neighbourhood of $x_\lambda$, then $U_u\cap A_u=\emptyset,\:u\in\lambda\:$.
Therefore $\prod U _{\lambda}\cap\prod A_\lambda=\prod(U_\lambda\cap A_\lambda)=\emptyset $(The product of the empty set is the empty set). So $(\prod A_\lambda)^c$ contains $U=\prod U _{\lambda}$ that is the neighbourhood of an arbitrary $x$. So $(\prod A_\lambda)^c$ is open which implies $(\prod A_\lambda)$ is closed.
I was told I should use $\prod \overline{A_\lambda}$ and $\overline{\prod A_\lambda}$. But I am not seeing how can I prove using the adherence.
Questions:
1)Is my proof right
2)How can I integrate $\prod \overline{A_\lambda}$ and $\overline{\prod A_\lambda}$ in my proof?
If my proof is wrong. Can someone provide one proof of the statement?
Thanks in advance!
 A: My first intuition of a proof was the following:
Let's denote the index set by $\Lambda$ and fix $\lambda_0 \in \Lambda$.
$A_{\lambda_0} \subset X_{\lambda_0}$ is closed, so $X_{\lambda_0} \setminus A_{\lambda_0} \subset X_{\lambda_0}$ is open. 
By definition of the product topology on $X := \prod\limits_{\lambda \in \Lambda} X_{\lambda}$, the set $(X_{\lambda_0} \setminus A_{\lambda_0}) \times \prod\limits_{\lambda \neq \lambda_0} X_{\lambda}$ is open in $X$.
So its complement $A_{\lambda_0} \times \prod\limits_{\lambda \neq \lambda_0} X_{\lambda}$ is closed in $X$.
The set $\prod\limits_{\lambda_0 \in \Lambda} A_{\lambda_0} = \bigcap\limits_{\lambda_0 \in \Lambda} \left( A_{\lambda_0} \times \prod\limits_{\lambda \neq \lambda_0} X_{\lambda} \right)$ is closed in $X$ as an intersection of closed sets.
Using closures will essentially lead to the same argument.
Edit: As mentioned in the comments, the openness of $(X_{\lambda_0} \setminus A_{\lambda_0}) \times \prod\limits_{\lambda \neq \lambda_0} X_{\lambda} = \pi_{\lambda_0}^{-1} \left( X_{\lambda_0} \setminus A_{\lambda_0} \right)$ comes from the fact, that the product topology is defined by requiring the projections onto the factors to be continuous: The product topology is the smallest (coarsest) topology containing all preimages of open sets under projections.
Warning: Infinite products of open sets are not open in general.
A: All $\pi_\lambda$ are continuous maps. So inverse images of closed sets are closed.
So $$\prod_{\lambda \in \Lambda} A_\lambda = \bigcap_{\lambda \in \Lambda} (\pi_\lambda)^{-1}[A_\lambda]$$
which is an intersection of closed sets so closed.
