Let $\mathcal A$ be the set of all Borel subset $B$ of $\Bbb R$ such that $f^{-1}(B)$ is also a Borel subset of $\Bbb R$. Since $f$ is Borel-measurable we have $(c,\infty)\in \mathcal A$ for all $c\in\Bbb R$.
Let $\sigma(\mathcal A)$ be the smallest $\sigma$-algebra containing the set $\mathcal A$. Since, the operation $f^{-1}$, i.e. operation of taking inverse commutes with the countable union operation and taking complement operation, so we have $\sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)=\big\{f^{-1}(X): X\in\sigma(\mathcal A)\big\}.$
Now, since $\sigma(\mathcal A)$ is a $\sigma$-algebra we have $(a,\infty)\cap (b,\infty)=(a,b)\in \sigma(\mathcal A)$ for all $a,b\in\Bbb R$.
Similarly, $(-\infty,a']=\Bbb R\backslash (a',\infty)$ is also in $\sigma(\mathcal A)$ for all $a'\in\Bbb R$ as $\sigma$-algebra is closed under complement.
Hence, $(-\infty,a)=\bigcup_{n=1}^\infty\big(-\infty,a-\frac{1}{n}\big]$ is also an element of $\sigma(\mathcal A)$ for all $a\in\Bbb R$ as $\sigma$-algebra is closed under countable union.
Also, every open subset of $\Bbb R$ can be written as a countable union of open intervals of $\Bbb R$ and every $\sigma$-algebra is closed under countable union.
Therefore, every open subset of $\Bbb R$ is an element of $\sigma(\mathcal A)$. In other words, the set $\tau(\Bbb R)$ of all open subsets of $\Bbb R$ is a subset of $\mathcal A$.
But, the Borel-$\sigma$ algebra $\mathcal B(\Bbb R)$ of $\Bbb R$ is the smallest $\sigma$-algebra containing all open subsets of $\Bbb R$, i.e. $\sigma\big(\tau(\Bbb R)\big)=\mathcal B(\Bbb R)$. Hence, $\sigma(\mathcal A)\supseteq \mathcal B(\Bbb R)$ as $\mathcal A\supseteq \tau(\Bbb R)$.
Finally, For any $Y\in\mathcal B(\Bbb R)\implies Y\in \sigma(\mathcal A)\implies f^{-1}(Y)\in \sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)\subseteq \mathcal B(\Bbb R)$. The last inclusion is due the fact that each set $f^{-1}(B)\in \mathcal B(\Bbb R)$ for all $B\in \mathcal A$ from definition of $\mathcal A$. Hence, $\sigma\big(\{f^{-1}(B):B\in\mathcal A\}\big)\subseteq \sigma\big(\mathcal B(\Bbb R)\big)=\mathcal B(\Bbb R)$.
If $A$ is a subset of $\Bbb R$ of measure $0$, then all its subsets, $2^{|A|}$ many of them, are measure $0$ (and so Lebesgue measurable, Borel or not). Any subspace of a separable metric space is itself separable metrisable and so has at most $\mathfrak{c} = |\Bbb R|$ many Borel subsets.
So whenever $2^{|A|} > \mathfrak{c}$, $A$ has a Lebesgue measurable subset that is not Borel in $A$. This is the case for the standard Cantor middle third set e.g.
Best Answer
Here are some hints:
To see why $E^{x_1}$ is borel, consider the (obviously borel) function $f(x_2) = (x_1,x_2)$. Do you see why $E^{x_1}$ is the preimage of a borel set by a borel function?
Remember that the lebesgue measurable sets are exactly $\{ \text{borel sets} \} \cup \{ \text{extra null sets} \}$, so we will probably have to use these extra null sets somehow.
More concretely: Take your favorite (lebesgue) nonmeasurable set $X$. Can you show that $X \times \{0\}$ is (lebesgue) measurable in $\mathbb{R}^2$? Remember you'll probably have to use nullsets! Once you've done this, $(X \times \{0\})^0$ is obviously $X$, which was nonmeasurable by assumption.
I hope this helps ^_^