A nice proof of Bolzano-Weierstrass

Following is a very nice proof for Bolzano-Weierstrass that I discovered. I think that any proof of B-W would require an infinite construction, and this one is no exception--However I feel that the infinite construction in this proof is easier to the intuition than the constructions in other proofs.
The theorem:

If is a sequence of numbers in the closed segment , then it has a subsequence which converges to a point in .

Let's take an arbitrary point

, which is somewhere between the points

and

. Observe the segment

. It may contain a finite number of members from the sequence

and it may contain an infinite number of them. If we were to take the point

to be

, the segment

would obviously contain an infinite number of members from the sequence. If we take

to be

, the segment

would contain at most only one point from the sequence. Introducing the set

Sidenote: A segment containing zero elements from

will be considered to contain a finite number of them.

We know that

belongs to

. There is one more thing we need to note about the set

. If a point

belongs to

, that would mean that

has a finite number of members from

in it, and that would therefore mean that any subset of

would also have only a finite number of members from

. Hence for any

that belongs to

, all the points between that

and

would also belong to

. We can see that

is actually a segment, starting at

and ending in some unknown location in

. Now we make our next move:

We are now going to show that

is an accumulation point of

.
Let's ignore the special case

, and assume that

. Now we take an arbitrarily small

. Observe the segment

cannot belong to

since it is higher than the supremum. Hence

contains an infinite number of

members. Now observe the segment

must belong to

, since it is smaller than the supremum of the segment

. Thus

contains a finite number of members from

. But

is a subset of

. If the bigger set contains an infinite number of

members and its subset contains only a finite amount, the complement of the subset must contain an infinite number of members from

. We have proved that for every

, the segment

contains an infinite number of members from the sequence. To rephrase that, we have proven that there exists a point

in the segment

, such that every neighborhood of

includes an infinite number of members from the sequence. We have proven that

is an accumulation point of

, from which it is easy to prove that there exists a subsequence of

which converges to

, and that spells Q.E.D.

In the special case that

it can still be shown that

is an accumulation point. I will not expand upon that since it is too boring.

For the benefit of the novice I will explain in detail how to prove that an accumulation point of a sequence is a limit of one of its subsequences. We already noted that for any

contains an infinite number of members from

. Now we will construct a subsequence of

that converges to

. Take

to be

. Take any

member in

to be our first member. Take

to be

. Take an

member in

with an index number higher than the previous member to be our next member--You are guaranteed to find one since there are an infinite number of

members in that segment. Take

to be

, and now find an

member in

with an index number higher than that of our previous member. Again you are guaranteed to find one. This process can continue indefinitely, thus constructing a subsequence of

that converges to

. (I told you an infinite construction has to come at some point!)