X is defined as $\{a,b,c,d\}$
A topological space $X$ is called $T_{0}$ if for every pair of points $x,y \in X$, there is an open set $U$ that contains one of them and not the other. Is $XT_{0}$ is under your topology (which you must construct) which is neither discrete nor trivial, and to prove it.
For the first part, I defined my topology to be $\{\emptyset, X, \{a\},\{ab\},\{abc\}\}$ because this topology is neither discrete nor trivial. However, how can I show that $XT_{0}$ is under my topology?
Indeed, $\mathcal{T} = \{\emptyset, X, \{a\}, \{a,b\}, \{a,b,c\} \}$ is a topology on $X = \{a,b,c,d\}$ because it forms a finite "chain" containing $X$ and $\emptyset$. For any two open subsets $U,V$ we have $U \subseteq V$ or $V \subseteq U$. This implies being closed under finite intersections: any finite set of open sets has a minimum one in terms of inclusion and this is the intersection, and the same holds for unions as all unions are finite and have a maximum set as well.
That it is $T_0$ is also clear in this instance:if $x \neq y$ then one of them, say $x$, is the "letter" with the lowest alphabetic position, and then $U$, being the set of all letters before and including $x$ is open and does not contain the other point.
Cleaner is to define $X = \{0,1,2,3\}$ and define $$\mathcal{T} = \{\emptyset\} \cup \{ U_j:=\{i: i \le j\} : j \in X\}$$
which comes down to the same thing essentially. Then $i \in U_i, j \notin U_i$ whenever $i < j$, etc.