I am trying to prove:
Prove that every finite group $G$ is isomorphic to a subgroup of the symmetric group $S_n$ for some $n$.
Here is my attempt, which is my attempt at trying to understand a proof in a set of algebra lecture notes I found.
For each $g \in G$, define the map \begin{align*} m_g: \underset{x}{G} \underset{\longmapsto}{\to} \underset{gx}{G}. \end{align*} I claim that $m_g \in \mathrm{Perm}(G)$, i.e., that $m_G$ is a bijection of sets. Given $a,b \in G$ for which $m_g (a) = m_g (b)$, we have $ga = gb$, so by left cancellation, we have $a = b$, so $m_g$ is injective. Furthermore, given $h \in G$, we have $m_g\left(g^{-1} h\right) = g\left(g^{-1} h\right) = h$, so $m_g$ is surjective and hence bijective. Next, define the map \begin{align*} \varphi: \underset{g}{G} \underset{\longmapsto}{\to} \underset{m_g}{\mathrm{Perm}(G)}, \end{align*} First, we show that $\varphi$ is a homomorphism, i.e., that given $g,h \in G$, we have $\varphi(gh) = \varphi(g) \varphi(h)$, which is equivalent to proving that $m_{gh} = m_g m_h$. For any $x \in G$, we have \begin{align*} m_{gh} (x) = (gh)x = g(hx) = m_g (hx) = m_g \left(m_h (x)\right) = m_g \circ m_h (x), \end{align*}
So $\varphi$ is a group homomorphism, as desired. Next, we show that $\varphi$ is injective. Given $g, g' \in G$ for which $g \neq g'$, we have \begin{align*} m_g (e) = ge = g \neq g' = g'e = m_{g'} (e), \end{align*} so $\varphi(g) \neq \varphi(g')$. This implies that $G \cong \mathrm{Im}(\varphi)$. Indeed, the map $f: G \to \mathrm{Im}(\varphi)$ sending $g \to \varphi(g)$ gives an isomorphism of groups. If $h \in \mathrm{Im}(\varphi$, then there exists a unique $g \in G$ such that $\varphi(g) = h$, so we have $f(g) = \varphi(g) = h$, so $f$ is surjective. Given $a,b \in G$ for which $f(a) = f(b)$, we have $\varphi(a) = \varphi(b)$, which implies that $a = b$ by injectivity of $\varphi$, so $f$ is injective and hence bijective. Finally, if $a,b \in G$, we have \begin{align*} f(ab) = \varphi(ab) = \varphi(a) \varphi(b) = f(a) f(b), \end{align*} since $\varphi$ is a group homomorphism. Clearly, $\mathrm{Im}(\varphi) \subset \mathrm{Perm}(G)$. Furthermore, $\mathrm{Perm}(G) \cong S_n$. Therefore, we conclude \begin{align*} G \cong \mathrm{Im}(\varphi) \subset \mathrm{Perm}(G) \cong S_n, \end{align*} as desired.
There are two things I have doubts about.
First, I haven't proved that $G$ is ismorphism to a subgroup of $S_n$, but to a subgroup of $\mathrm{Perm}(G)$. Since $\mathrm{Perm}(G)$ and $S_n$ are isomorphic, I can view $\mathrm{Perm}(G)$ as a subgroup of $S_n$. Is this the correct idea? Alternatively, since $\mathrm{Perm}(G) \to S_n$, there exists an isomorphism $\Phi$ between the two, and if $\mathrm{Im}(\varphi)$ is a subgroup of $\mathrm{Perm}(G)$, then the image of this subgroup under the ismorphism $\Phi$ is a subgroup of $S_n$.
Second, I can't manage to prove that $\mathrm{Perm}(G)$ and $S_n$ are isomorphic. In particular, I'm having trouble understanding the differences between them. $\mathrm{Perm}(G)$ is the set of maps $f: G \to G$ that are bijective and $S_n$ is the set of permutations of $\{1, \ldots, n\}$. If I label the elements of $G$ as $\{x_1, \ldots, x_n\}$, then these are clearly the same set, up to labelling, but I don't know how to prove that is the case.
Concerning your first question, the answer is affirmative: since $\operatorname{Perm}(G)$ and $S_n$ are isomorphic, you can view $\operatorname{Perm}(G)$ as a subgroup of $S_n$.
On the the other hand, given any two sets $A$ and $B$ for which there is a bejection $\psi\colon A\longrightarrow B$, the groups $\operatorname{Perm}(A)$ and $\operatorname{Perm}(B)$; just take$$\begin{array}{rccc}\varphi\colon&\operatorname{Perm}(A)&\longrightarrow&\operatorname{Perm}(B)\\&\eta&\mapsto&\psi\circ\eta\circ\psi^{-1}.\end{array}$$In particular, since $S_n=\operatorname{Perm}(\{1,2,\ldots,n\})$, if $G$ has $n$ elements, then $\operatorname{Perm}(G)$ and $S_n$ are isomorphic.