If there is a finite subset of $T$ with empty intersection, then any filter that contains $T$ would necessarily contain the empty set (since filters are closed under finite intersections), and this is impossible. So the given condition is certainly necessary.
To see that the given condition is sufficient, consider the collection $U$ of all subsets of $S$ that contain a finite intersection of elements of $T$ (the empty intersection being defined to be $S$ itself). I claim that this is a filter on $S$.
Certainly, $U$ contains $S$, since $S$ contains every element of $T$. Since no finite intersection of elements of $T$ is empty, the empty set does not contain any finite intersection of elements of $T$, so $\varnothing\notin U$. If $A$ contains $T_1\cap\cdots \cap T_n$ and $B$ contains $R_1\cap\cdots\cap R_m$, where $T_i,R_j\in T$, then $A\cap B$ contains $(T_1\cap\cdots\cap T_n)\cap(R_1\cap\cdots\cap R_m)$, which is a finite intersection of elements of $T$; hence $A\cap B\in U$. And if $A\in U$ then $A$ contains a finite intersection of elements of elements of $T$, and so does any $B\subseteq S$ that has $A\subseteq B$. So if $A\in U$ and $A\subseteq B$, then $B\in U$. Thus, $U$ is a filter on $S$, as claimed. Moreover, $S$ contains $T$.
Finally, I claim that if $\{U_i\}$ is a nonempty family of filters on $S$, each of which contains $T$, then $\bigcap U_i$ is a filter on $S$ that contains $T$. Indeed, let $U$ denote the intersection. Since $S\in U_i$ for all $i$, then $S\in U$. Since $\varnothing\notin U_i$ for each $i$, $\varnothing\notin U$. If $A,B\in U$, then $A,B\in U_i$ for each $i$, so $A\cap B\in U_i$ for each $i$, hence $A\cap B\in U$. If $A\in U$ and $A\subseteq B$, then $A\in U_i$ for each $i$, hence $B\in U_i$ for each $i$, hence $B\in U$. And finally, since $T\subseteq U_i$ for each $i$, then $T\subseteq \bigcap U_i = U$.
So now simply consider $\bigcap \{ U \mid U\text{is a filter on }S\text{ that contains }T\}.$ We know the family is nonempty, so the intersection makes sense; it is a filter that contains $T$; and by construction it is the smallest filter on $S$ that contains $T$. Thus, the condition is sufficient.
(In fact, the specific example that was constructed above is the smallest filter that contains $T$, since any filter that contains $T$ must contain every set that contains all finite intersections of elements of $T$.)
You'll note that the one example I constructed was constructed as "the filter of all sets that contain some finite intersection of elements of $T$". That means the filter whose base are the "finite intersection of elements of $T$".