$\forall n\in\mathbb N$, let $A_n=\{k\in\mathbb N\mid k\geq n\}$. Determine if the following statement is true or false: $\displaystyle\bigcap_{k\in\mathbb N}A_k=\emptyset$. Justify your conclusion.
I'm quite sure the statement is true. Here is my proof. Any feedback would be appreciated greatly:
Proof. We will prove this by contradiction. So suppose, for the sake of a contradiction, that $\displaystyle\bigcap_{k\in\mathbb N}A_k\neq\emptyset$ and let $a\in\bigcap_{k\in\mathbb N}A_k$. This means that $\forall m\in\mathbb N$, $a\in A_m$, which implies that $\forall m\in\mathbb N$, $a\geq m$.
Let $b=a+1$. Observe that $b\in\mathbb N$ since $\mathbb N$ is closed under addition and $b\gt a$. This contradicts the fact that $a\geq m$ for all natural numbers $m$. $$\tag*{$\blacksquare$}$$
Your proof is OK, but I'd write it differently:
Suppose we have some $m \in \bigcap_{k \in \mathbb{N}} A_k$. Then $m$ is in all $A_k$, in particular $m \in A_{m+1}$ giving $m \ge m+1$, a flagrant contradiction with $m < m+1$. So we cannot have such $m$ and the intersection is empty.