The following says that subsequences converge to the same thing that a convergent sequence converges to. We leave the proof to you.
The following is the definition of a Cauchy sequence in ℝ^{p}. I assume the reader is familiar with Cauchy sequences in ℝ or ℂ. Completeness of these is the same as saying that all Cauchy sequences converge.
Definition 2.2.19

One has the same set of Cauchy sequences if
A sequence is Cauchy, means the terms are “bunching up to each other” as m,n get large.
Theorem 2.2.20 The set of terms in a Cauchy sequence in ℝ^{p} is bounded in the sense that for all n,
Proof: Let ε = 1 in the definition of a Cauchy sequence and let n > n_{1}. Then from the definition,

It follows that for all n > n_{1},

Therefore, for all n,

Proof: Let ε > 0 be given and suppose a_{n}→ a. Then from the definition of convergence, there exists n_{ε} such that if n > n_{ε}, it follows that

Therefore, if m,n ≥ n_{ε} + 1, it follows that

showing that, since ε > 0 is arbitrary,
Suppose then that

Therefore, the sequence converges. ■
This is not unique to ℝ^{p}. The same holds for any normed linear space.
There is no change replacing
Proof: Let
As mentioned above, sequential compactness and compactness are equivalent in ℝ^{p}. The following is a very important property pertaining to compact sets. It is a surprising result.
Proposition 2.2.23 Suppose ℱ is a nonempty collection of nonempty compact sets with the finite intersection property. This means that the intersection of any finite subset of ℱ is nonempty. Then ∩ℱ≠∅.
Proof: First I show each compact set is closed. Let K be a nonempty compact set and suppose p
Consider now the claim about the intersection. If this were not so,

and so, in particular, picking some F_{0} ∈ℱ,

would be an open cover of F_{0}. A point in F_{0} is not in F_{0}^{C} so it must be in one of the above sets. Since F_{0} is compact, some finite subcover, F_{1}^{C},

which means ∩_{k=0}^{m}F_{k} = ∅, contrary to the finite intersection property. ■
Note that absolutely no mention was made of context. This is because this finite intersection property is always true whenever you have a set of compact sets.