Are you assuming your group $G$ is finite? In this case we can proceed by induction on $|G|$. You should add in your proof "Since $G$ is solvable, there is a [normal] subgroup $H$ such that both $H$ and $G/H$ are solvable [and $1<|H|<|G|$]." You have three cases:
-- if $H\subseteq K$, by inductive hypothesis, $K/H$ is a solvable subgroup of $G/H$ (where $|G/H|<|G|$ since $1<|H|$), and the fact that both $H$ and $K/H$ are solvable gives $K$ solvable;
-- if, $K\subseteq H$, then by inductive hypothesis $K$ is solvable (use $|H|<|G|$);
-- in the remaining case, you have $|K\cap H|<|H|$, so, by inductive hypothesis $K\cap H$ is solvable. Furthermore, $K/(K\cap H)\cong KH/H$ is a subgroup of $G/H$ and it is therefore solvable (use that $|G/H|<|G|$).
The "standard" proof:
Consider, for each $i$, the subgroups $G_iN/N$ of $G/N$. It is straight-forward to show that $G_iN$ is normal in $G_{i+1}N$ (and thus $G_iN/N$ is normal in $G_{i+1}N/N$).
Now $\dfrac{G_{i+1}N/N}{G_iN/N} \cong G_{i+1}N/G_iN$.
Taking any $x,y \in G_{i+1}$, and $n,n' \in N$, we see that:
$[xn(G_iN),yn'(G_iN)] = [xn,yn']G_iN$, and since $N \lhd G$, $[xn,yn'] \in [x,y]N$, so that:
$[xn,yn']G_iN = [xn,yn']NG_i = ([xn,yn']N)G_i = $
$([x,y]N)G_i = [x,y]NG_i = [x,y]G_iN = ([x,y]G_i)N$.
Since $G_{i+1}/G_i$ is abelian, $[x,y]G_i = G_i$ so $[xn,yn']G_iN = G_iN$,so that
$G_{i+1}N/G_iN$ is abelian, as desired.
(here, $[x,y] = xyx^{-1}y^{-1}$ and we are using the fact that for a group $G$, $G$ is abelian iff $[x,y] = e$ for all $x,y \in G$).
It should be noted that some of the quotients $G_{i+1}N/G_iN$ may be trivial, resulting in a shorter subnormal series for $G/N$.
Best Answer
Hint: $G/(K \cap N)$ can be homomorphically embedded in $G/K \times G/N$. And use that a product of solvable groups is again solvable and that subgroups of solvable groups are again solvable.