From a Young diagram we can build an irreducible complex representation for the symmetric group. It turns out that they are all dinite dimensional and that every other representation of $S_n$ is isomorphic to one constructed via a Young diagram.
So in particular, there are really no infinite dimensional irreducible complex representations of $S_n$?
On
You can prove stronger statement: every complex irreducible representation of a compact group is finite dimensional. Since finite group is compact, the statement for finite group follows. Here is a good reference for the subject: https://en.wikipedia.org/wiki/Compact_group
Perhaps I should make my comment into an answer.
Let $\rho$ be any representation of any finite group $G$ over any field $K$, and suppose that the representation is acting on the (nonzero) vector space $V$.
Choose any nonzero vector $v \in V$. Then it is straightforward to check that the subspace of $G$ spanned by the vectors $\{ \rho(g)(v) : g \in G \}$ is invariant under the action of $G$ (that's because $G$ permutes the elements of the spanning set). So if the representation is irreducible then it has degree at most $|G|$.
You can go a little further, because if this space has dimension exactly $|G|$, then the spanning set forms a basis, and it has a proper $G$-invariant subspace of codimension $1$, consisting of those vectors for which the coefficient sum is $0$. So any irreducible representation of $G$ has degree at most $|G|-1$.
This bound is attained for example for cyclic groups of prime order $p$ over finite fields of order $q$ such that the multiplicative order of $q$ mod $p$ is $p-1$.