There is a nice and not too long way to decompose $Sym^2\mathbb{C}[X]$ (permutation representation of $S_5$ acting on $\{x_1,\cdots,x_5\}$) in irreducible without using characters?
I know a way using characters, since it is sufficient to know the characters of the standard and trivial representation (using $\chi_{sym^2V}(g) = \frac{1}{2}\chi_V(g^2)+\chi_V(g)^2)$ and then taking the scalar product with all irreducible characters of $S_5$ watching at the table characters of $S_5$.
I'm wondering whether there's a way without using characters (schur functions or symmetric polynomial together with Schur functors are allowed)
Any help or tip would be appreciated.
Yes, this can be done as follows, and this argument generalizes to $S_n$ with no difficulty.
$S^2$ of a permutation representation is itself a permutation representation: namely it's the permutation representation on the set of $2$-element multisets of $X$, which you can think of as $S^2(X)$. This permutation representation decomposes into two orbits, the multisets $\{ x, x \}$ and the subsets $\{ x, y \}$ where $x \neq y$. So we get
$$S^2(\mathbb{C}[X]) \cong \mathbb{C}[X] \oplus \mathbb{C} \left[ {X \choose 2} \right]$$
where ${N \choose 2}$ is the set of $2$-element subsets. To finish from here use the result from this math.SE answer which tells us that each representation $\mathbb{C} \left[ {X \choose k} \right]$ embeds into the next one $\mathbb{C} \left[ {X \choose k+1} \right]$ with irreducible cokernel. This still uses characters but it does not involve knowing the character table.
The conclusion is that $S^2(\mathbb{C}[X])$ decomposes into two copies of $1$, two copies of a $4$-dimensional irreducible, and a $5$-dimensional irreducible.