Proof that $S_3$ and $S_4$ are solvable groups

First of all let me point out that $D_{24}$ cannot be isomorphic to $S_4$. $S_4$ does not have an element of order 12, where as $D_{24}$ does. This is because $D_{24}$ as a set is

$$\{1,r,r^2 ,\ldots r^{11}, sr, sr^2, \ldots sr^{11}\}$$

where the $r_i's$ are the rotations and the $sr^j$ the reflections; so that $r$ is your element of order 12. For the sake of knowledge though $S_4$ does have a subgroup of order 12 sitting inside it that is just $A_4$. Just to tell you though the subgroups of $S_4$ are (up to isomorphism):

$$\{e\}, \hspace{2mm} C_2, \hspace{2mm} C_3, \hspace{2mm} V_4, \hspace{2mm} C_4, \hspace{2mm} S_3, \hspace{2mm} D_8, \hspace{2mm} A_4, \hspace{2mm} S_4.$$

Actually it is not so hard to see directly that $S_3$ and $D_6$ are isomorphic, just list down the elements explicitly and if they obey exactly the same relations, their Cayley Tables are the same so you have your isomorphism.

By the way I think for your last line you want a surjective homomorphism from $S_4$ to $S_3$ to show that $S_3$ is solvable only if $S_4$ is. The way you do that is like this:

We get a homomorphism $\varphi$ from $S_4$ to $S_3$ by considering the action of $S_4$ on a finite set $Y = \{\Pi_1, \Pi_2, \Pi_3\}$ where $\Pi_1 = \left\{\{1,2\}, \{3,4\}\right\}$, $\Pi_2 = \left\{\{1,3\}, \{2,4\}\right\}$ and $\Pi_3 =\left\{ \{1,4\}, \{2,3\}\right\}$. Each $\Pi_i$ is one way of partitioning a set of 4 elements into two subsets of equal cardinality. $S_4$ acts on a $\Pi_i$ by permuting the numbers in each partition, so for example the cycle $(1234)$ interchanges $\Pi_1$ and $\Pi_3$ while keeping $\Pi_2$ fixed, so that $\varphi( (1234) ) = (13)(2)$. It is easy to see that $\ker \varphi$ is the Klein four- group $V_4= \{e, (12)(34), (14)(23), (13)(24)\}$ where $e$ denotes the identity in $S_4$. If you go further you should know that by the first isomorphism theorem the quotient group $S_4/V_4$ is isomorphic to the image of $\varphi$.

Now to see the surjectivity of $\varphi$, recall we have the counting formula

$|S_4| = |\ker \varphi||\operatorname{Im} \varphi|$. Therefore the order of the image is $24/4 = 6$. Now recall the image is a subgroup of $S_3$ so if we have a subgroup of order 6 sitting inside $S_3$, it must be the whole of $S_3$.

To answer your question on why $S_4$ is solvable, note that the only normal subgroups in $S_4$ are the trivial group, the Klein 4-group $V_4$, $A_4$ and $S_4$ itself.

So we have a subnormal series

$$\{e\} \triangleleft V_4 \triangleleft A_4 \triangleleft S_4.$$

Now you should know that $V_4/\{e\} \cong V_4$ that is abelian (all groups of order 4 are). You also know that $|A_4/V_4| = 3$ so that $A_4/V_4 \cong C_3$ which is abelian. Finally $|S_4/A_4| = 2$ so that $S_4/A_4 \cong C_2$ that is also abelian. So it remains to check that $A_4$ is normal in $S_4$ and $V_4$ is normal in $A_4$. The former you already know for the index of $A_4$ in $S_4$ is two. For the second I suggest listing out the conjugacy classes of $A_4$ and show that $V_4$ is a union of conjugacy classes. It should not be so hard to compute all the conjugacy classes of $A_4$ since there are only 12 elements. Once you've done all of this, you've proved that $S_4$ is solvable!

That $D_6$ is isomorphic to $S_3$ is easy: consider how $D_6$ acts one the corners of the triangle of which it is the symmetry group. However $D_{24}$ is not isomorphic to $S_4$ (consider elements of order $12$, or even of order $6$, which $S_4$ does not posess), so no amount of effort will suffice to prove it that way. The injective homomorphism (one of the many) $S_3\to S_4$ will not help you with $S_4$ either. However there is a surjective homomorphism $S_4\to S_3$ (think of a tetrahedron and the lines joining the midpoints of its sides) that will help you to prove that $S_4$ is solvable.