Isn't the given characterisation of recursively enumerable subsets of the class of all recursively enumerable languages?
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 $B$.