$S_3$ acts on $\lbrace 1,2,3 \rbrace$, so this affords a homomorphism $S_3\to GL_3(\mathbb{C})$ (acting on $\mathbb{C}^3$). I showed the only vector fixed by the action of $S_3$ is zero.
Find two different vector subspaces $U,V$ preserved by the image of $S_3$ such that $\mathbb{C}^3=U\oplus V$.
How can $U,V$ be constructed? Since $U$ has to be fixed by permutations of its three components, its elements must be of the form $(a+bi,a+bi,a+bi)$, and the same can be said for $V$. Based on this fact, I cannot see how $\mathbb{C}^3=U+V$. For example, $(0,1,0)$ will not belong to either $U$ or $V$.
There's some misunderstanding in the requirement of the task: You are required to find subspaces that are as sets stable under the permutation action, but not necessarily element-wise.
Concerning your comment on $(0,1,0)$ neither belonging to $U$ nor $V$: note that ${\mathbb C}=U\oplus V$ does not imply that any $v\in {\mathbb C}^3$ belongs to either $U$ or $V$, but only that it can be written as a linear combination of elements of $U$ and $V$.