If $B$ is infinite, there is a injective function $\beta:\mathbb{N} \to B$.
Since $A$ is finite, there is some $N$ such that for $n > N$, $\beta(n) \notin A$.
Then the function $\beta'$ defined by $\beta'(n) = \beta(n+N)$
is an injective function $\beta':\mathbb{N} \to B\setminus A$,
and so $B\setminus A$ is infinite.
Here is a way to prove it directly.
If $B$ is infinite, there is a injective function $\beta:\mathbb{N} \to B$.
Since $A$ is finite, there is some $N$ such that for $n > N$, $\beta(n) \notin A$.
Then the function $\beta'$ defined by $\beta'(n) = \beta(n+N)$ is an injective function $\beta':\mathbb{N} \to B\setminus A$, and so $B\setminus A$ is infinite.