Why is the following statement impossible for a sequence $\{p_n\}$? 
A sequence is said to be convergent if there is a point to which it converges. A convergent sequence cannot converge to two distinct limits. For if $\{p_n\}$ were to converge to both $p$ and $q$ with $p \ne q$, then we could choose spherical neighborhoods about $p$ and $q$ that are disjoint; but, all but a finite number of the $p_n$ would have to be inside each of these neighborhoods, which is impossible.

I'm a little confused by the following statement.

but, all but a finite number of the $p_n$ would have to be inside each of these neighborhoods, which is impossible.

Is it impossible by definition of a convergent sequence?

Let $(x_n)$ be a sequence converging to distinct $x$ and $y$. Since $x$ and $y$ are distinct, there are $\varepsilon,\delta>0$ such that $B_\varepsilon(x)\cap B_\delta(y)=\varnothing$. By definition of convergence, there are $M$ and $N$ such that $|x_m-x|<\varepsilon$ and $|x_n-y|<\delta$ whenever $m>M$ and $n>N$. However, pick an integer $p>\max\{M,N\}$. Then this says that $x_p\in B_\varepsilon(x)\cap B_\delta(y)=\varnothing$, which makes no sense.
