If a set $A$ is finite then $A\cap B$ is a finite set. Background: 
Theorem - If $A\subseteq \mathbb{N}_n$ then $A$ is a finite set and $|A|\leq n$.
Question:

Let $A$ be a finite set and $B$ be some set. If $A$ is a finite set, then $A\cap B$ is a finite set.

Attempted proof:
Let $A$ be a finite set and let $C = A\cap B$. If $C = \emptyset$ then $C$ is finite. Suppose $C \neq \emptyset$, since $C\subseteq A$, the set $A\neq \emptyset$ and there exists $k\in\mathbb{N}_k$ such that $A\sim \mathbb{N}_k$. That is, there exists $k\in\mathbb{N}$ and there exists a one-to-one correspondence $f:A\to\mathbb{N}_k$. The restriction of $f$ on the set $C$ $f|_C$ is a one-to-one function from $C$ onto $f(C)$. Therefore, $C\sim f(C)$. By theorem above, some $f(C)$ is a subset of $\mathbb{N}_k$, $f(C)$ is a finite set therefore since $C\sim f(C)$, $C$ is finite as well.
I am not sure if this is completely right, any feedback or other approaches would be appreciated.

Suppose $A\cap B$ is infinite. Then we have $x_1,x_2,x_3,....$ all in $A\cap B$. But then each of them is also in $A$. Then $A$ has ininite cardinality. A contradicition.
