A two-part problem:
$\bullet$ Let $H \leq S_n$ be the subgroup that fixes 1. Show that $H$ is isomorphic to $S_{n-1}$.
$\bullet$ Show that there are no proper subgroups of $S_n$ that properly contain $H$.
For the first part, can I argue that if 1 is fixed, then we are permuting $n-1$ objects, and so there are $(n-1)!$ elements in this group?
For the second part, I'm not sure. I tried using index arguments, but I don't think they work. I cannot see why, for example in $S_4$ which contains $H$ and $|H| = 6$, why an index 2 subgroup, say $A_4$ of order 12, or some index 3 subgroup of order 8 cannot contain $H$.
The first part is basically true. If the one is fixed, all the remaining permutations are exactly those on a $4$ element set. You can also easily write down this specific isomorphism $H \to S_{n-1}$.
For the second part I would argue as follows: Suppose there is a subgroup $K \leq S_n$ with $H \subsetneq K \subseteq S_n$. Then $K$ has to contain a permutation $\pi$ with $\pi(1) = c \neq 1$. Now using this $\pi$ and the $\sigma \in H$ can you build every permutation in $S_n$? If so we have $K=S_n$ which is what we want to show.
Edit: Some more details. Every permutation can be written as product of nontrivial disjoint cycles. In particular $\pi=\sigma_1 \circ \sigma_2 \circ \dots \circ \sigma_m$ can be written as product of disjoint cycles $\sigma_i$. As $\pi(1) = c \neq 1$ there must be a cycle $\sigma_i$ with $\sigma_i(1) = c$. Let $$ \sigma_i = (1 \; c \; c_3 \; \cdots \; c_k )$$ be this cycle. Now define $$\sigma = \sigma_1^{-1} \circ \dots \circ \sigma_{i-1}^{-1} \circ (c_k \; c_{k-1} \; \cdots \; c_3 \; c) \circ \sigma_{i+1}^{-1} \circ \dots \circ \sigma_m^{-1} \in H \subseteq K. $$ Then $\sigma \circ \pi = (1 \; a_k) \in K$ with $a_k \neq 1$, so we have our first transposition in $K$. From this transposition we can get every transposition: Let $b \neq 1$, then $$ (\ast \; \ast) (1 \; a_k) (\ast \; \ast) = (1 \; b) $$ for some suitable transposition $(\ast \; \ast) \in H$. If you can find this transposition you know that all transpositions are in $K$. As every permutation can be written as product of transpositions we must have $K=S_n$.
Edit 2: A concrete example using $S_5$ (in smaller groups certain arguments simplify a bit). We know that there is an $\pi \in K$ which does not fix $1$. The permutation $\pi$ is the product of disjoint cycles. Suppose $$ \pi = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 1 & 5 & 4\end{pmatrix} = (1 \; 2 \; 3) (4 \; 5). $$ Now we define (as in the "algorithm" above): $$ \sigma = (4 \; 5) (3 \; 2).$$ This $\sigma$ has $1$ fixed, so $\sigma \in H \subseteq K$. You can easily check that $\sigma \circ \pi = (1 \; 3)$ and as $K$ is a group it is closed under multiplication, i.e. $(1 \; 3) \in K$. This is the first transposition. How do we get all the others? We use $$ (b \; 3)(1 \; 3)(b \; 3) = (1 \; b) $$ for $b \neq 1$ and have $(b \; 3) \in K$, so $(1 \; b) \in K$ for all $b$. The other transpositions (not involving $1$) are in $K$ anyways as $H \subseteq K$. So $K$ contains all transpositions and from these you can build every permutation. Hence $K = S_5$.