f is measurable iff its coordinate functions are measurable 
I am really struggling to connect the sets in $\mathcal{B}(\mathbb{R^n})$ and $\mathcal{B}(\mathbb{R})$. 
Both inclusions are causing me problems. This questions seems a lot harder than it looks.
 A: Consider for every $i=1,\dots,n$ the projection
$$ \pi_i\colon \mathbb{R}^n \to \mathbb{R} $$
onto the $i$-th factor.
By definition, the $\sigma$-algebra $\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra on $\mathbb{R}^n$ such that every projection $\pi_i$ is measurable (w.r.t. the $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ on $\mathcal{B}(\mathbb{R}))$.
First we show that a map
$$ f\colon (E,\mathcal{E}) \to (\mathbb{R}^n,\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R})) $$
is measurable if and only if every projection $f_i$ is measurable.By the very definition, the projections
$$ \pi_i \colon (\mathbb{R}^n,\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R}))\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
are measurable, so that, if $f$ is measurable then every $f_i=\pi_i\circ f$ is measurable as well, because composition of measurable functions is measurable.
Conversely, suppose that every $f_i$ is measurable. By definition, we can see that $\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R})$ is generated by the family of sets
$$ \mathcal{D} =  \left\{ B_1\times \dots \times B_n \,|\, B_i\in \mathcal{B}(\mathbb{R}) \right\}=\left\{ \pi_1^{-1}(B_1)\cap \dots \cap \pi_n^{-1}(B_n)  \,|\, B_i\in \mathcal{B}(\mathbb{R}) \right\}$$ 
so that it is enough to show that $f^{-1}(D)\in \mathcal{E}$ for every $D\in \mathcal{D}$. However we see that 
$$ f^{-1}(\pi_1^{-1}(B_1)\cap \dots \cap \pi_n^{-1}(B_n))=f_1^{-1}(B_1)\cap \dots \cap f_n^{-1}(B_n) $$
and since all the $f_i$ are measurable we are done.

To conclude, it is enough to show that
$$ \mathcal{B}(\mathbb{R}^n) = \mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R}) $$
It is easy to see that $\mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R})$ is generated by the family of sets
$$ \mathcal{A}=\left\{ A_1\times \dots \times A_n \,|\, A_i\subseteq \mathbb{R} \text{ open } \right\} $$
and as each one of the sets in $\mathcal{A}$ is open in $\mathbb{R}^n$ it follows that
$$ \mathcal{B}(\mathbb{R}^n)\supseteq \mathcal{B}(\mathbb{R})\otimes \dots \otimes \mathcal{B}(\mathbb{R}) $$
For the other inclusion, it is enough to show that every open subset of $\mathbb{R}^n$ is a countable union of subsets in $\mathcal{A}$. Basically we want to prove that every open subsets is a countable union of small open hypercubes. To prove this formally one could proceed like this: consider the norm on $\mathbb{R}^n$ given by
$$ |\cdot |_1 \colon \mathbb{R}^n \to \mathbb{R} \qquad (x_1,\dots,x_n) \mapsto \max{|x_i|}  $$
then the balls for this norm are precisely small open hypercubes: indeed for every $y=(y_1,\dots,y_n)\in \mathbb{R}^n$ we have
$$ B_1(y,\epsilon)=\left\{ x \in \mathbb{R}^n \,|\, |x-y|_1 < \epsilon \right\} = \left(y_1-\frac{\epsilon}{2},y_1+\frac{\epsilon}{2}\right)\times \dots \times \left(y_n-\frac{\epsilon}{2},y_n+\frac{\epsilon}{2}\right)  $$
Now, every two norms on $\mathbb{R}^n$ are equivalent, and in particular the norm $|\cdot |_1$ defines the same topology of the standard euclidean norm, i.e. the euclidean topology on $\mathbb{R}^n$. Now, one can show that the set
$$ \mathcal{U}= \left\{ B_1(y,\epsilon) \,|\, y\in \mathbb{Q}^n, \epsilon \in \mathbb{Q}_{\geq 0} \right\} $$
is a countable basis of open sets, so that every open subset of $\mathbb{R}^n$ is a countable union of subsets that belong to $\mathcal{A}$.
