Is that possible to partition $(\Bbb R,+)$ into 4 additively closed subsets?

You can partition $(\Bbb R^n,+)$ into the three subsemigroups $H^n_+=\{x\in\Bbb R^n\,:\, x_n>0\}$, $H^n_-=\{x\in\Bbb R^n\,:\,x_n<0\}$ and $\Bbb R^{n-1}\times\{0\}$. Therefore, if $\Bbb R^{n-1}$ can be partitioned into $k$ subsemigroups, then $\Bbb R^n$ can be partitioned into $k+2$ subsemigroups. Now, let $m$ be the smallest positive natural number such that $(\Bbb R,+)$ cannot be partitioned in $m$ subsemigroups. As you said, $m\ge3$, and by definition $(\Bbb R,+)$ can be partitioned in $m-2$ subsemigroups. But then $(\Bbb R^2,+)$ can be partitioned in $m$ subsemigroups. But thanks to AC (axiom of choice), $(\Bbb R,+)$ and $(\Bbb R^2,+)$ are isomorphic as groups, and therefore $(\Bbb R,+)$ can be partitioned into $m$ subsemgroups as well, against the hypothesis on $m$. Therefore $(\Bbb R,+)$ can be partitioned into any finite number of subsemigroups.