What is the smallest possible basis for the finite complement topology on $\mathbb{R}$?

Let $\mathscr{B}_0=\{U\in\mathscr{T}:|\Bbb R\setminus U|\text{ is even}\}$ and $\mathscr{B}_1=\{U\in\mathscr{T}:|\Bbb R\setminus U|\text{ is odd}\}$. $\mathscr{B}_0$ and $\mathscr{B}_1$ are both bases for $\mathscr{T}$, but $\mathscr{B}_0\cap\mathscr{B}_1=\varnothing$, so $\mathscr{T}$ has no base that is a subset of both $\mathscr{B}_0$ and $\mathscr{B}_1$ and therefore no smallest base.

No, a non-discrete T$_1$ topology cannot have a minimal base.

For suppose $\mathscr{B}$ is a base for such a topology. Then we can choose $B \in \mathscr{B}$ with $|B|\gt 1$, since a topology with a base consisting only of singleton sets is discrete.

We show that $\mathscr{B}$ is not minimal by showing that $$ \mathscr{B}^\prime \colon= \mathscr{B} \setminus \{B\} $$ is a base for the topology, and we show that by showing that $B$ is a union of sets in $\mathscr{B}^\prime$.

Consider any point $x \in B$; we have to find a set $B^\prime \in \mathscr{B}^\prime$ such that $$ x\in B^\prime \subseteq B.$$

Since $|B|\gt1$, we can choose a point $y\in B$ such that $y \neq x$. As the topology is T$_1$, so there is an open set $U$ such that $x \in U$ and $y \not\in U$.

Since $\mathscr{B}$ is a base, there is a set $B^\prime \in \mathscr{B}$ such that $$ x \in B^\prime \subseteq B\cap U.$$ Since $y \not\in U$, therefore $y \not\in B \cap U$ either and thus $y \notin B^\prime$, but as $y \in B$, so we have $B^\prime \neq B$. Thus $B^\prime \in \mathscr{B}$ and $B^\prime \neq B$; that is, $$ B^\prime \in \mathscr{B}^\prime.$$

In particular, the cofinite topology on an infinite set is a non-discrete T$_1$ topology, so it has no minimal base.