How to show that every $A ∈ L(V)$ has a unique decomposition as a sum of a symmetric and an antisymmetric operator? I need to show that every $A ∈ L(V)$ has a unique decomposition as a sum of a symmetric and an antisymmetric operator, where $V$ be an Euclidean vector space with scalar product $(.|.)$.
A linear map $A ∈ L(V)$ is said to be symmetric (real-self-adjoint), if
$$A^T = A,$$
it is said to be antisymmetric (scew-symmetric), if
$$A^T = −A.$$
What if I represent $A$ as $\dfrac {A+A^T}{2}+\dfrac {A-A^T}{2}$?

You already wrote $A$ as a sum of symmetric and anti-symmetric operator so you answered the existence question. Now assume that you can write 
$$A= S+M=S'+M'$$
where $S$ and $S'$ are symmetric and $M$ and $M'$ are anti-symmetric so
$$S-S'=M'-M\implies S=S'\;\text{and}\; M=M'\quad \text{why ?}$$ 
