The first part is too complicated: $A\subseteq AB$ (whether or not $AB$ is a group), and $A\cap B\subseteq A$; so you are done as far as inclusion; and intersection of subgroups is always a subgroup, so $A\cap B$ is a subgroup and is contained in $AB$. Of course, you should note that $AB$ actually is a subgroup, because $AB=BA$ by the normality of $A$.
Your final assertion does not follow. You have justified correctly that $k'\in A$; but you have not said a single word as to why it is in $B$. So how do you justify your assertion that $k'\in A\cap B$?
The reason it is in $B$ is that $k\in A\cap B$, so $k\in B$. Therefore, $bkb^{-1}$, being a product of elements in $B$, must be in $B$. Since it is also in $A$ (by the reasons given), you h ave that $k'=bkb^{-1}\in A\cap B$.
Using some suggestions from the other commenters:
The alternating group, $A_4$, has the set $H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4$ as a subgroup. If $f\in S_4\supseteq A_4$ is a permutation, then $f^{-1}[(12)(34)]f$ has the effect of swapping $f(1)$ with $f(2)$ and $f(3)$ with $f(4)$. One of these is $1$, and depending on which it is paired with, the conjugated element may be any of $H-\{I\}$, since the other two are also swapped. Thus $H\lhd S_4$ is normal, so $H\lhd A_4$ as well. Similarly, $H$ has three nontrivial subgroups, and taking $K=\{I,(12)(34)\}\cong C_2$, this is normal because $V_4$ is abelian. But $K\not\lhd A_4$, since $$[(123)][(12)(34)][(132)]=(13)(24)\in H-K.$$
Moreover, this is a minimal counterexample, since $|A_4|=12=2\cdot 2\cdot 3$ is the next smallest number which factors into three integers, which is required for $K\lhd H\lhd G$ but $\{I\}\subset K\subset H\subset G$ so that $[G\,:\,H]>1$, $[H\,:\,K]>1$, $|K|>1$ and
$$|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.$$
The smallest integer satisfying this requirement is 8, but the only non-abelian groups with $|G|=8$ are the dihedral group $D_4$ and the quaternion group $Q_8$, and neither of these have counterexamples. (Note that if $G$ is abelian, then all subgroups are normal.) Thus $A_4$ is a minimal counterexample. (Edit: Oops, $D_4$ has a counterexample, as mentioned in the comments: $\langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4$, but $\langle s\rangle\not\lhd D_4$.)
However, if $H\lhd G$ and $K$ is a characteristic subgroup of $H$, then $K$ is normal in $G$. This is because the group action $f$ defines an automorphism on $G$, $\varphi(g)=f^{-1}gf$, and because $H$ is normal, $\varphi(H)=H$ so that $\varphi|_H$ is an automorphism on $H$. Thus $\varphi(K)=K$ since $K$ is characteristic on $H$ and so $\{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G$.
Best Answer
You know that $H\unlhd G$ and $K\unlhd G$. This means that for any $g\in G$ we have $ghg^{-1}\in H$ and $gkg^{-1}\in K$, where $h\in H$ and $k\in K$. So take any $x\in H\cap K$, since $x\in H$ we have $gxg^{-1}\in H$, by the same argument we have $gxg^{-1}\in K$, so $gxg^{-1}\in H\cap K$. Thus $(H\cap K)\unlhd G$.