Let $a_n \rightarrow A$ and $b_n \rightarrow B$ as $n \rightarrow \infty$, with $a_n \leq b_n$ for all $n$. Show $A \leq B$. This is my proof:
We have the sequence $a_n$ is convergent. Consider some arbitrary $b_k$. We have for all $n$ that $a_n \leq b_k$. By the Limit Location theorem, we have $\lim_{n\rightarrow \infty} a_n \leq b_k$, i.e. $A \leq b_k$. 
Since $b_k$ is arbitrary, we can generalize this inequality to all $b_n$: namely, for all $n$, $b_n \geq A$. By the Limit Location Theorem, $\lim_{n \rightarrow \infty} b_n \geq A$, i.e. $B \geq A$. QED. 
Is this correct?

No: we don't know that $a_n\leq b_k$ for all $n$.  We only know that $a_n\leq b_n$ (in other words, that $a_n\leq b_k$ when $n=k$).
