Calling $\emptyset$ and $A$ "improper" subsets of a set $A$ is not universal, and it is confusing in this case, because the meaning of "proper" is not the same. It is standard to say that $S$ is a proper subset of $A$ if (and only if) every element of $S$ is an element of $A$, but $S$ is not equal to $A$, i.e., at least one element of $A$ is not in $S$. Under this definition, $\emptyset$ is a proper subset of every nonempty set, even though it is "improper" according to the convention you were also given. Just remember that mathematical terminology varies and isn't always logical. Here "improper" does not mean "not proper". (For this reason I would personally not use the convention of calling the empty set "improper".)
When finding all proper subsets, you should count the empty set.
Let us forget temporarily about the "proper" subset restriction on $B$. So we want to divide $X$ into $4$ pairwise disjoint parts, $A_1,A_2,A_3,A_4$, and then let $A=A_1$, $B=A_1\cup A_2$, and $C=A_1\cup A_2\cup A_3$.
Given such a partition of $X$, there is an associated function $f:X\to\{1,2,3,4\}$ given by $f(x)=i$ if and only if $x\in A_1$. And given any function $f:X\to \{1,2,3,4\}$, we can use $f$ to produce a partition of $X$ into sets $A_1,A_2, A_3,A_4$, and then produce uniquely determined sets $A\subseteq B\subseteq C\subseteq X$.
There are $4^n$ functions from $X$ to $\{1,2,3,4\}$. So without the proper subset restriction, there are $4^n$ ways to do the job. We leave dealing with the restriction to you. From $4^n$ we must subtract the number of ways to choose $A,B,C$ such that $A\subseteq B=C\subseteq X$.
Best Answer
Consider the complement.
Hint: There are $2^n$ pairs of subsets where $A = B$, which is a subset of $S$.
This follows from the rule of product, because there are 2 possibilities for each element.
Hint: There are $3^n$ pairs of subsets where $A$ is a subset of $B$, which is a subset of $S$.
This follows from the rule of product, because there are 3 possibilities for each element.