A confusion regarding an infinite sequence. Sorry this question might be unclear/coherent, but it baffles me logically. 
Say we have an infinite sequence $\{x_i\}\forall i\in\Bbb{N}$. Can we choose every point in the sequence? Say we start from $x_1$, and choose every point after that; i.e. $x_2,x_3,x_4,\dots$. Remember that we're choosing the points individually, and not selecting the whole set $\{x_i\}$. Can we choose every point in $\{x_i\}$?
Say we choose a point $x_i$. Then there are infinite points $x_j$ left in $\{x_i\}$, $j>i$. This is true regardless of however high the value of $i$ is. I suppose the point $\lim _{n\to\infty}x_n$ can't be selected. 
Let us suppose that we cannot select every point in $\{x_i\}$. When we say property $A$ is true for $\{x_i\}$, we prove it by saying "select any $x_i\in \{x_i\}$", and then prove it for $x_i$. Does this mean that there are point in $\{x_i\}$ for which property $A$ might not be true, as those points can't be selected in such a fashion?
Thanks in advance! An exposition on this would be greatly appreciated. 
 A: Let $A$ be a property, and let's consider it as a Boolean function on the points of the sequence $\{x_n\}$.  That is $A(x_n)=$ true if $x_n$ has property $A$, and $A(x_n)=$ false if $x_n$ does not have property $A$.  Suppose also we can prove the implication:
$$\left(x_i\in\{x_n\}\right)\Longrightarrow (A(x_i)=\text{ true})$$
Notice this implication states that property $A$ holds for all values of the sequence $x_i$.  In proving this is true, we didn't have to select every element individually, we just prove it for a general representative $x_i$.  As long as we don't assume anything special about $x_i$, except that it is a member of $\{x_n\}$, then the above implication shows that $x_i$ has property $A$.
As a simple example, let $x_1=1$, $x_2=2$, $x_3=3$, $\ldots$, $x_n=n$, $\ldots$.  Notice that if $x_i\in\{x_n\}$, then $x_i=i>0$.  It follows that every member of the sequence is positive, because every member is of the form $x_i$ for some $i$, and we just showed it is true for arbitrary $i$.
Edit (to address comments about convergence):
Let's play a game with the sequence $x_n=\frac{1}{n}$.  First, you choose a point in the sequence, and then I try to choose a ball around $0$ not containing this point.  If I can do this, I win; if not, I lose.  Who wins this game?  As you've shown in the comments, I do.
Now, let's try another game.  I'll first pick a ball around $0$, and then you try to exhibit a value $N$ such that $x_n$ is always in this ball as long as $n>N$.  If you can do this, you win, otherwise you lose.  You'll quickly see that you should be able to win this game, precisely because the sequence converges to $0$.
Notice the first game shows that no single point of the sequence is contained in every ball around $0$, and the second game shows that any ball around $0$ contains infinitely many points of the sequence.  These two statements are not contradictory.
A: @Ayush, I'd rather choose another wording to address your concern in your last comment below your question: given 
$$x_i\in\{x_n\}_{n\in\Bbb N}\;\;\text{and}\;\;l:=\lim_{n\to\infty}x_n\;,\;\;\text{let}\;\;\epsilon:=|l-x_i|$$
and assume $\,x_i\ne l\implies \epsilon>0\;$ , then we get that
$$x_i\notin\left\{x\in\Bbb R\;;\;|l-x|<\frac\epsilon2\right\}=:B\left(l\,,\,\frac\epsilon2\right)$$
and thus we can say that in any neighborhood of the limit $\;l\;$ there exist elements of the sequence that aren't contained in it...but this, of course, does not contradict the definition of limit which implies that  any neighborhood of the purposed limit must contain all the elements of the sequence except perhaps a finite number of them ...which is precisely what happens here.
