Proving $\otimes_{i=1}^{i=n}\mathcal{B}_{X_{i}}=\mathcal{B}_{\Pi X_{i}}$ I am given the following exercise:

Let $X_{\alpha}$ be a measureable space with $\sigma$-algebra
  $M_{\alpha}$ , mark $X\triangleq{\displaystyle \prod_{\alpha\in
 A}X_{\alpha}}$ and $\pi_{\alpha}:X\rightarrow X_{\alpha}$ .
define $\otimes_{\alpha\in A}M_{\alpha}$ as the $\sigma$-algebra that
  is created from sets of form
  $\pi_{\alpha}^{-1}\left(E_{\alpha}\right)$ where $E_{\alpha}\in
 M_{\alpha}$

Part 1 of the question asked to prove that if $A$ is countable then $\otimes_{\alpha}M_{\alpha}$
is created by sets of form ${\displaystyle \prod_{\alpha\in A}E_{\alpha}}$
where $E_{\alpha}\in M_{\alpha}$ .
Given a topological space $X$ denote the Borel's $\sigma$ -algebra
on $X$ by $\mathcal{B}_{X}$
Consider $X_{1},\dots,X_{n}$ metric spaces and define $X=X_{1}\times\dots \times X_{n}$
Now the question asks:

a) Prove
  $\otimes_{i=1}^{i=n}\mathcal{B}_{X_{i}}\subseteq\mathcal{B}_{X}$.
b) Assume each $X_{i}$ is separable and prove
  $\otimes_{i=1}^{i=n}\mathcal{B}_{X_{i}}=\mathcal{B}_{X}$.

I am having some problems with this question, mainly because I don't
have a strong memory of what the product topology is exactly (I have
a definition near me and so I am trying to work with that).
For $a$ I want to start with some element of the form $\Pi_{\alpha}E_{\alpha}$and
prove that it is an element of $\mathcal{B}_{x}$ but to be honest,
I am not completely sure how $\mathcal{B}_{X}$ looks like - how does
the open set in $X$ look like and why something of the form $\Pi_{\alpha}E_{\alpha}$
(or $\pi^{-1}(E_{\alpha})$) is such a set. 
After I understand $a$ I hope I will have an idea for $b$
I would really appreciate the help!
 A: I will give a sketch of proof (it's homework!) in the case $n=2$; it's not hardly adaptable to the general case. 
a) For $i\in\{1,2\}$ and $B_i\in\mathcal{B}_{X_i}$, we have $\pi_i^{-1}(B_i)\in\mathcal B_X$, as $\pi_i\colon \mathcal B_X\to \mathcal B_{X_i}$ is assumed to be measurable. Now, to get the result, we use the fact that $\mathcal B_X$ is a $\sigma$-algebra containing the sets of the form $\pi_i^{-1}(B_i), i\in\{1,2\}, B_i\in\mathcal B_{X_i}$. 
b) We have to show that $\mathcal{B}_{X_1}\otimes \mathcal{B}_{X_2}$ contains the open subsets of $X_1\times X_2$. It's not hard to see it for open sets of the form $O_1\times O_2$, where $O_i$ is open in $X_i$. Let $\{a_n\}$ and $\{b_n\}$ sequences respectively dense in $X_1$ and $X_2$, and $O\subset X_1\times X_2$ open. Fix $(x_1,x_2)\in O$. We can find open sets $O_i$ such that $(x,y)\in O_1\times O_2\subset O$. If $X_i$ has a countable base of open sets, we are done. 
If the space are assumed to be metric, then separability implies that $X_i$ has a countable basis of open sets. But it's not true in general, as tomasz pointed out. 
Note that separability is needed, otherwise take $X_1=X_2=$ a "big set" with discrete topology. 
A: For (a) note that the product topology on $\prod_{i=1}^n X_i$ is generated by "open cubes", that is sets of the form $\prod_{i=1}^n U_i$ where each $U_i \subseteq X_i$ is open. As $n=2$ suffices by induction. Let now $U \subseteq X_1$ open and look at $\{V \subseteq X_2 \mid U \times V \in \mathcal B_X\}$. By the open, this contains all open sets. Now show it's a $\sigma$-algebra. Now let $V \in \mathcal B_{X_2}$ and consider $\{ U\subseteq X_1 \mid U \times V \in \mathcal B_X\}$. We know that this contains all open $U$. Now conclude.
