I would appreciate if you could please express your opinion on this proof. The proof appears to be "self-evident", but, as it often happens with one trying to gain more algebraic maturity, it may not be rigorous enough, especially when it comes to apparently the most simple of things.
Proof:
$S_X$ is a group of permutations on the set $X$, thus $|S_X| = |X|! = n!$. This implies that a bijection exists between $S_X$ and $S_n$. Hence, $S_X\cong S_n$.
Let me try to spell out what has been said in the comments.
Showing that two groups are isomorphic involves exhibiting an isomorphism between them. As noted in the comments, you need to identify the groups involved: Let $X$ be a set. The group $S_X$ is the set of all bijections $X\to X$ with multiplication given by composition of functions.
Now, let $X$ and $Y$ be two sets of order $n$. It is not enough that $|S_X|=|S_Y|$ since an arbitrary bijection between the two need not be a homomorphism. You need to produce a bijection that is.
Try this: Let $\gamma:X\to Y$ be any bijection. Define a map $\Phi_\gamma:S_X\to S_Y$ by $$\Phi_\gamma(\sigma)=\gamma\circ\sigma\circ\gamma^{-1}$$ You need to show that $\Phi_{\gamma}(\sigma\circ\tau)=\Phi_\gamma(\sigma)\circ\Phi_\gamma(\tau)$ and that $\Phi_\gamma$ is bijective (hint: show that $\Phi_\gamma^{-1}=\Phi_{\gamma^{-1}}$).