I have a problem that saying
Suppose $K$ is a subgroup normal of $G$, and $H$ is a subgroup of $G$. Show that $K \cap H$ is a normal subgroup of $H$.
I've tried to show it by first showing that $K \cap H$ is not the empty set, because identity element is in $K \cap H$. Then, I show that $K \cap H$ is a subgroup. After that, I use the normal subgroup test:
Take arbitrary element $a \in K \cap H$, this means that b is element of K and element of H.
$aha^{-1} = h1hh2 = h$ is element of $H$, for all a element of $K \cap H$, and $h$ element of $H$
So, $K \cap H$ is normal subgroup of $H$. However, I'm not really sure about this, and I wondering what is the use of the information saying that $K$ is normal subgroup of $G$?
Evidently we know that $K \cap H$ is a subgroup of $H$ given that the intersection of subgroups is a subgroup and $K \cap H \subset H$. As you ascertained then, the only thing left to establish is that $K \cap H$ is normal in $H$.
Normalcy boils down to the condition that the each left coset of $K \cap H$ is equivalent to the right coset generated by the same element (that is, $h(K \cap H) = (K \cap H)h$, where $h \in H$). We can easily establish this is equivalent to the following condition:
$$\forall h \in H, \forall a \in H \cap K, hah^{-1} \in H \cap K $$
Note then what normalcy looks like for $K \subset G$; we have:
$$\forall g \in G, \forall k \in K, gkg^{-1} \in K$$
Let then for this sake $h \in H$ and $a \in H \cap K$. As $H \subset G$ and $H \cap K \subset K$, we notice $hah^{-1} \in K$ by the previous condition. Moreover, $h, a,$ and $h^{-1}$ are in $H$ by assumption, so $hah^{-1} \in H$. Combining these yields $hah^{-1} \in H \cap K$, and so $H \cap K$ is normal in $H$.
I think your confusion may stem from an error in your definition of normalcy. If we want for example $K$ normal in $G$, we want to show the conjugation of any element of $K$ by an element in $G$ (i.e. $gkg^{-1}$) is an element of $K$, not just an element of $G$ (as this follows by closure).