Isn't the given characterisation of recursively enumerable subsets of the class of all recursively enumerable languages? $S$ is a subset of the class of all recursively enumerable languages over some finite symbols then $S$ is recursively enumerable iff


If $L$ is in $S$ and $L'$ is a language such that $L ⊆ L'$ and $L'$ is recursively enumerable, then $L'$ is in $S$
If $L$ is an infinite language in $S$, then there exists at least one finite subset of $L$ that is in $S$
The set of all finite languages in $S$ is enumerable, i.e. a Turing machine can list all the finite languages in $S$


Source of the statement: https://cs.stackexchange.com/q/2322 and some online notes

Isn't the 3rd one contradictory to the 1st one?
No, it is not contradictory. Both conditions are positive in the sense that they "put things in $S$", so they cannot be at odds with each other.

There is an equivalent formulation of $S$: there is an r.e. set $B$ of finite languages such that
$$S = \{L \mid \text{$L$ is r.e. and $\exists L_0 \in B \,.\, L_0 \subseteq L$}\}.$$
In words: $S$ is generated by an r.e. set of finite language $B$ in the following way: $S$ contains precisely all r.e. languages which contain an element of $B$.