...where the addition and scalar multiplication is defined as $A+B=\{a+b|a\in A \ and \ b\in B\}$ and $\alpha . A = \{\alpha a | a \in A\}$, where $\alpha \in \mathbb{R}$.
I think this claim is false, since $A=[0,1]$ and $-A=[-1,0]$ are convex subsets of $\mathbb{R}$ but $A+(-A)\neq {0}$. Is my reasoning correct?
Is it possible to create a vector space with intervals or non-discrete sets?
Thank you in advance for any tip and advice.
Your reasoning is correct, though you might first want to verify that the zero element with respect to this addition is the nonempty convex subset $\{0\}\subset\Bbb{R}$.
Then it is indeed the case that $$A+(-A)=\{a+a':\ a\in[0,1],\ a'\in[-1,0]\}\neq\{0\},$$ because $1+0=1$ and $0+(-1)=-1$ are elements of $A+(-A)$, but not of $\{0\}$.
Edit: As noted in the comments, you might also want to verify that no other nonempty convex subset of $\Bbb{R}$ is the inverse of $A$, i.e. no other nonempty $B\subset\Bbb{R}$ satisfies $$A+B=B+A=\{0\}.$$ Of course $A\subset A+B$ so all convex nonempty subsets of $\Bbb{R}$ fail to have an inverse, except $\{0\}$ itself.
Alternatively, if the set is a vector space then multiplication distributes over addition and so $$A+(-A)=1\cdot A+(-1\cdot A)=(1+(-1))\cdot A=0\cdot A=\{0\},$$ which we have just shown to be false.