Compactness in a metric space

**Problem**:

Let X = the set of real number sequence such that

Define

Let

Let **N**, let

Define , is compact?

**My proof**:

First, we claim that there exist a subsequence of every sequence in that converges in X.

Suppose that is a sequence in , note that

Now,

Note that , so is bounded. By a theorem, we know that there exist a subsequence of such that converges to a point in **R**, denote by .

Now, consider the component in the subsequence , which is also bounded. Thus there exist a subsequence of such that converges to a point in **R**, this time denote by .

Note that the component is a subsequence of , thus it also converges to the point .

Continue in the manner, we find a subsequence of , denote by for simplification purpose, such that **N**.

Now, consider . Thus proves that .

Now, we claim that is a point in .

Assume to the contrary that is not a point in . We have

Note that since it is a sequence in .

But, , in which contradicts our assumption. Therefore we show that , thus proves is compact.

**Q.E.D.**

Please check my proof, thank you!