I know one way to construct a Bernstein set that is an additive group.Here is the way that I know. $\{P_\xi\colon \xi<\mathfrak c\}$ all nonempty perfect subsets of $\Bbb R.$ Choose, by recursion on $\xi<\mathfrak c$, sequence $\{x_\xi\colon \xi<\mathfrak c\}$ and $\{y_\xi\colon \xi<\mathfrak c\}$ so that $$ x_\xi,y_\xi\in P_\xi\setminus\Bbb Q\bigg(\{x_\zeta\colon \zeta<\xi\}\cup \{y_\zeta\colon \zeta<\xi\}\bigg ).$$ This choice can be made since
$$\Bigg|\Bbb Q\bigg(\{x_\zeta\colon \zeta<\xi\}\cup\{y_\zeta\colon \zeta<\xi\}\bigg)\Bigg|<\mathfrak c$$ Now we define $B=\{x_\xi\colon \xi<\mathfrak c\}.$ Notice that $ B$ is Bernstein set. In addition, $B$ is linear independent over $\mathbb Q.$
Next, let $B_1$ be linear space generated by $B.$ So, $B_1$ is as needed, that for any perfect $P\subset\mathbb R$ we have $$P\cap B_1\neq\emptyset\ \ \ \ \text{and} \ \ P\setminus B_1\neq\emptyset$$
Indeed, $B_1\cap P\neq\emptyset$ since $B$ does so. Consider the quotient group $\mathbb R\setminus B_1,$ for contradiction, assume $P\subset B_1.$ Now, let $x+B_1$ be a nonzero coset and where $x\notin B_1$ so $x+P\subset x+B_1.$ Notice that $x+P$ is a perfect set and $(x+P)\cap B_1=\emptyset$ which is a contradiction.
I was wondering if there is another way to construct $B_1.$ I hope from those that have a deep understanding and more knowledge to share their ideas. So, student like me they will get a lot of benefits from that. Any hint or help will be appreciated greatly.