Conditions for "transitivity" of normality I'm wondering if the problem is the following problem is as straightforward as it seems:

Problem: Let $H \leq K \leq G$.  If $H \unlhd G$, then $H \unlhd K$.
Solution: Since $H \unlhd G$, we know that $gHg^{-1}=H$.  Since $K \leq G$ and $H \leq K$, we have that $kHk^{-1}=H$.  Hence $K \unlhd H$.

 A: The proof is as easy as you expected, but the term for the property is not.
For normality to be a transitive relationship, we would require that $H\unlhd K$ and $K\unlhd G$ imply that $H\unlhd G$, but this is known to be false for groups in general.
In a poset (like the poset of subgroups of a group $G$), a family of elements of the poset is said to be "downward closed" if whenever $A\leq B$ and $B\in S$, then also $A\in S$. There is, of course, a similar definition for "upward closed". Naturally, no finite intervals $(a,b)$ are either upward or downward closed.
To generalize and rephase what you said slightly: $\{K\leq G\mid H\unlhd K\}$ is closed downward in the poset $\{K\leq G \mid H\leq K\}$". Actually $\{K\leq G\mid H\unlhd K\}$ has a maximum element: $N_G(H)$.
More examples
Any interval of the real line of the form $(-\infty, a)$ is of course a downward closed set under the ordinary ordering of the real line, and likewise $(a,\infty)$ is upward closed.
As another example, for any set $X$ with powerset $\mathcal{P}(X)$, let $\mathcal{F}$ denote the finite subsets of $X$. Then $\mathcal{F}$ is closed downward in $\mathcal{P}(X)$. Likewise, let $\mathcal{I}$ denote the subset of $\mathcal{P}(X)$ with all the infinite subsets of $X$. Then $\mathcal{I}$ is closed upward in $\mathcal{P}(X)$.
Somewhat similarly to groups, you can show that the poset of subrngs does not have a transitive "ideal" relation. However, just as with the group example, the subrngs containing a subrng $S$ which have $S$ as an ideal are closed downward in the poset of subrngs containing $S$.
I'm trying (and failing) to come up with simpler examples of upward closed and downward closed sets . The only examples that keep occurring to me are a little advanced.
Essential submodules of a module form an upward closed set (any submodule containing an essential submodule is itself essential.)
Superfluous submodules (on the same wiki page) form a closed downward set of the submodules of a module (any submodule contained in a superfluous submodule is itself superfluous.)
In topology, any subset of $X$ containing a dense subset is itself a dense subset of $X$. So, the set of dense subsets of a topological space is closed upward.
