Can a finite subbasis create an infinite topology / topological space? I've translated the following from my German textbook, so please correct me if there is something wrong or strange.
Definitions
Let $X$ be a set and $\mathfrak{T} \subseteq \mathcal{P}(X)$ with fits the following restrictions:
(i)   $\emptyset, X \in \mathfrak{T}$
 (ii)  $\forall U_1, U_2 \in \mathfrak{T}: U_1 \cap U_2 \in \mathfrak{T}$
 (iii) Let $I$ be an index set with $\forall i \in I: U_i \in \mathfrak{T} $. Then: $\bigcup_{i \in I} U_i \in \mathfrak{T}$
Then $\mathfrak{T}$ is called a topology and $(X, \mathfrak{T})$ is called a topological space

Let $(X, \mathfrak{T})$ be a topological space. $\mathcal{S} \subseteq \mathfrak{T}$ is called a subbasis of $\mathfrak{T} : \Leftrightarrow \forall U \in \mathfrak{T}: U$ is a union of finite intersections from Elements in $\mathcal{S}$.
Questions
Can a finite subbasis generate an infinite topology? Or, more formally, is the following implication true:
$$|\mathcal{S}| \in \mathbb{N} \Rightarrow |\mathfrak{T}| \in \mathbb{N}$$
Can a finite subbasis generate any topology for an infinite space $X$? So, formally:
$$|\mathcal{S}| \in \mathbb{N} \Rightarrow |X| \in \mathbb{N}$$
 A: No.  If $\mathcal{S}$ is a finite, set, then there are only finitely many (nonempty) finite subcollections of that set, and so the family $\mathcal{I}$ of intersections of (nonempty) finite subcollections of $\mathcal{S}$ is also finite (saying "finite" here is somewhat redundant).  As there are only finitely many subcollections of $\mathcal{I}$, there are only finitely many distinct unions of subcollections of $\mathcal{I}$.  Thus any topology generated by a finite subbasis is itself finite.
In particular, the usual metric topology on $\mathcal{R}$ has no finite subbase.
(Even more, given an arbitrary subbasis $\mathcal{S}$ on a set $X$, either $\mathcal{S}$ and $\mathcal{I}$ (as described above) above are both finite, or are both infinite and of the same cardinality. It follows that the topology generated by $\mathcal{S}$ has cardinality bounded above by some cardinal related to $| \mathcal{S} |$.  In the finite case, it would be $2^{2^{|\mathcal{S}|}}$, and in the infinite case it is $2^{|\mathcal{S}|}$.)
