Standard representation of $S_n$ is irreducible Let $S_n$ act on an $n$-dimensional $\mathbb{Q}$-vector $V$ space with basis $\{v_1,\cdots,v_n\}$ by $\sigma(v_i)=v_{\sigma(i)}$. Consider subspaces 
$W_1=\langle v_1+v_2+\cdots + v_n\rangle$ and $W_2=\langle v_i-v_j : i\neq j\rangle$. Clearly $W_1$ is subspace with trivial action of $S_n$. My aim is to show that $W_2$ is irreducible subspace of $V$. 
For this, first note that $W_1\cap W_2=0$ and $W_1+W_2=V$. 
Next, suppose $U$ is a non-zero $S_n$-invariant subspace of $W_2$, and so take a vector $0\neq v\in U$. Then $v=\lambda_1v_1 + \cdots + \lambda_n v_n$ where not all $\lambda_i$ are equal. Without loss of generality, we can assume that $\lambda_1\neq \lambda_2$. 
Then for $g=(1,2)\in S_n$, $g.v\in U$ and so $g.v=\lambda_2v_1 + \lambda_1v_2+ \lambda_3v_3+ \cdots + \lambda_nv_n\in U$. Subtracting $g.v$ from $v$ we get that 
$$v-g.v=(\lambda_1-\lambda_2)(v_1-v_2)\in U \mbox{ and } \lambda_1\neq \lambda_2.$$
This implies that $v_1-v_2\in U$. From this, it is easy to deduce that $v_i-v_j\in U$ for all $i\neq j$. This forces that $U=W_2$. q.e.d.
Is this proof correct? 

I know that there are some proofs using 2-transitive action and/or character theory, but I am going by some different way, and want to see if it is correct.
 A: I see one problem.  How do you know that, for other indices $i\neq j$ apart from $(i,j)=(1,2)$, there is a nonzero $u=\sum_{k=1}^n\lambda_k v_k$ in $U$ such that $\lambda_i\neq \lambda_j$?  You can assume without loss of generality that, for one fixed pair $(i,j)$ (which you took to be $(1,2)$), such an element $u\in U$ exists.  But you start to lose "generality" when you assume further that other pairs $(i,j)$ can have this property too.  To clarify, what prevents $U$ from being spanned by $v_1-v_2$?  I would proceed in a different way.  
Since $v_1-v_2\in U$, as you correctly deduced, it follows that
$$(2\;j)\cdot(v_1-v_2)=v_1-v_j\in U$$
for $j=3,4,5,\ldots,n$.  That is, $v_1-v_2$, $v_1-v_3$, $\ldots$, $v_1-v_n$ are in $U$, but then the span of these elements are precisely $W_2$.
A: There is only one minor problem. It is not correct to assert that “From this, it is easy to deduce that $v_i-v_j\in U$ for all $i\neq j$.” What you should say is that by the same argument $v_i-v_j\in U$ for all $i\neq j$. Other than that, it's fine.
