With a classmate's permission, this is their solution to the problem (G is $A_n$ here):
Let $K$ be a field of characteristic $\neq 2$. Let $G$ be the alternating group $A_n$. Let $K\left[\boldsymbol{x}\right]$ be the polynomial ring $K\left[x_1, x_2, \ldots, x_n\right]$.
We claim that the invariant ring $K[\boldsymbol{x}]^G$ is generated by symmetric polynomials (so we may take the Schur polynomial basis) and the Vandermonde polynomial $P = \prod_{i < j} (x_j - x_i)$. Note that symmetric polynomials are invariant under $S_n$ (simply by definition) and also the Vandermonde polynomial is invariant under $A_n$ since an even number of transpositions swaps some two variables an even number of times, so the sign of $P$ also changes an even number of times and so it is invariant. We now show that these polynomials indeed generate the whole invariant ring.
It suffices to show that any polynomial $f$ that is invariant under the action of $A_n$ can be expressed as a $K[\boldsymbol{x}]^{S_n}$-linear combination of a symmetric polynomial and $P$. To do this, define $g(x_1, x_2, \dots, x_n) = f(x_2, x_1, x_3, \dots, x_n)$, i.e. $g$ is simply a polynomial $(1\;2)f$, the image of $f$ under the action of the transposition $(1\;2)$. In fact, it is irrelevant which odd permutation we pick in place of $(1\;2)$ while defining $g$, as any transposition can be written as a product of any other transposition and some even permutation (which we know fixes $f$ by assumption). Observe now that for any permutation $\pi\in S_n$ (and $\sigma$ any transposition), we get
$$\pi(f + g) = \pi f + \pi g = \pi f + \pi (\sigma f) = \pi f + \sigma' \pi f $$
where $\sigma'$ is a transposition. Note that if $\pi$ is even then $\pi (f + g) = f + g$, and similarly if $\pi$ is odd then $\pi (f + g) = g + f$, thus proving that $f + g$ is symmetric. Moreover, note that if $\pi$ is an odd permutation, we have $$\pi (f - g) = \pi f - \pi \sigma f = g - f$$ so $f-g$ changes sign when acted on by an odd permutation. This means that for any polynomial $f$ invariant under $A_n$, we may write
$$f = \frac{1}{2}(f+g) + \frac{1}{2}(f-g),$$ and so $f$ can be written as a $K[\boldsymbol{x}]^{S_n}$-linear combination of a symmetric polynomial and an alternating polynomial. It now suffices to show that $P$ divides any alternating polynomial to conclude the proof (since the quotient will be a symmetric polynomial).
Note that for $\sigma = (i\; j)$ and an alternating polynomial $h$, we have $\sigma h = - h$. Also, whenever $x_i = x_j$ we have $\sigma h(\boldsymbol{x}) = h(\boldsymbol{x})$. This means that $h(\boldsymbol{x}) = 0$ when $x_i = x_j$, and so $(x_i - x_j) \vert h$. This holds for any $i \neq j$ and the polynomial ring is a UFD, so we get that $P\vert h$, as desired.
Best Answer
If $n=3$ then there might be more subspaces, for example the span of $(1,\omega,\omega^2)$ where $\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$.
If $n>3$ then the subspaces you listed are indeed all the subspaces. Let $e_1,...,e_n$ be the standard basis vectors of $\mathbb{C^n}$. We can define an action of $A_n$ on $\{e_1,...,e_n\}$ by $\sigma.e_i=e_{\sigma(i)}$. Just like any other action, it induces a representation of $A_n$. Formally, the representation is defined as $\pi:A_n\to GL(\mathbb{C^n})$ where $\pi(\sigma)(a_1e_1+...+a_ne_n)=a_1e_{\sigma(1)}+...+a_ne_{\sigma(n)}$. So we are actually looking for the invariant subspaces of this representation.
We can obviously define the invariant subspaces $V_1$ and $V_2$ like you did in the question. The subspace $V_1$ is one dimensional, thus irreducible. Also, since $n>3$ the action of $A_n$ on $\{e_1,...,e_n\}$ is $2$-transitive. (can you prove it?). This implies that if $\chi$ is the character of $\pi$ and $\chi_1$ is the character of the trivial representation then $\chi-\chi_1$ is an irreducible character. Since $\mathbb{C^n}=V_1\oplus V_2$ it follows that $\chi-\chi_1$ is exactly the character of $V_2$, so $V_2$ is also irreducible.
Now the result follows. Take a non-trivial invariant subspace $U$. It has an invariant complement $W$, which also has to be non-trivial. By Maschke's theorem you can decompose both into a direct sum of irreducible subspaces. But such a decomposition of $\mathbb{C^n}$ is unique, it is exactly $V_1\oplus V_2$. So we have no choice, $U$ must be either $V_1$ or $V_2$.