In what order to prove different forms of the Bolzano-Weierstrass theorem and the Heine-Borel theorem is the easiest way? If I understand the Bolzano-Weierstrass theorem and the Heine-Borel theorem correctly,
1-3 are all statements of the Bolzano-Weierstrass theorem:

  
*
  
*Every bounded sequence in R has a convergent subsequence.
  
*A subset of R is sequentially compact iff it is closed and bounded.
  
*A subset of a metric space is sequentially compact iff it is complete and totally bounded.
  

4-5 are all statements of the Heine-Borel theorem:


  
*A subset of R is compact iff it is closed and bounded.
  
*A subset of a metric space is compact iff it is complete and totally bounded.
  

I think the B-W theorem is related to sequences and sequentially compact, and the H-B theorem is related to (covering) compact. There is another theorem stating that


  
*A subset of a metric space is compact iff it is sequentially compact.
  

But after reading several textbooks I think this theorem should have another name or just no name.
Question: In what order to prove 1-6 by oneself is the easiest way?
 A: Which order is easiest is mainly a matter of preference, so I'll restrict this answer to a possible path through these theorems. You write that you want to prove the statements by yourself, I'll state each of the theorems and some lemmas that will help you prove the theorems, and leave a sketch of the proofs in the spoilers. Try to prove each theorem and lemma without looking at the spoilers first, and use the spoilers if you get stuck. I've tried to keep the prrofs self-contained in the sense that they only refer to definitions and not to results, with the exception of the well-known fact that $\mathbb{R}$ is complete. 
We will start with the statements concerning general metric spaces, and after that specialize to the situation in $\mathbb{R}$. In all that follows, we let $(M, d)$ be some metric space and $X \subseteq M$ a subset (in fact the ambient space $M$ is not so relevant, but since you phrase the statements in terms of subsets of the metric space, I'll do the same). We start easy: 
Theorem 1. If $X$ is compact, then $X$ is totally bounded. 

 Let $\epsilon > 0$, and consider the collection of open balls $\mathcal{U} = \{B_\epsilon(x): x \in X\}$. By compactness, it has a finite subcover. 

Next up is completeness. This one is a bit harder, so we will start with two preliminary lemmas that will also be useful for the proof of theorem 3. (I'm not including a proof, if you get stuck let me know in the comments.) 
Lemma 1. Let $(x_n)$ be a sequence in $X$. Then $(x_n)$ is Cauchy if and only if for every $\epsilon > 0$ there is a closed ball $\overline{B}_\epsilon$ of radius $\epsilon$ containing all but finitely many of the elements of the sequence. 
Lemma 2. Let $(x_n)$ be Cauchy, and choose for every $\epsilon > 0$ a closed ball $\overline{B}_\epsilon$ as in the previous lemma. Then the sequence converges in $X$ if and only if $X \cap \bigcap_{\epsilon > 0} \overline B_\epsilon \neq \emptyset$. 
Using these lemmas, we can prove the following. 
Theorem 2. If $X$ is compact, then $X$ is complete.  

 We prove the contrapositive. Suppose $(x_n)$ is Cauchy but does not converge in $X$, and choose closed balls $\overline B_\epsilon$ as in lemma 2. Then the set $\mathcal{U} = \{B_\epsilon^c : \epsilon > 0\}$ is an open cover of $X$ by lemma 2. However, any finite subset of $\mathcal{U}$ misses infinitely many of the $x_n$, and in particular does not cover $X$. So $X$ is not compact. 

Now we now that compactness implies completeness and total boundedness, so we move the sequential compactness. We split the implication in two parts. The first is the following theorem. The proof is not very difficult but a bit tiresome to write down completely, so the spoiler contains only a sketch.  
Theorem. If $X$ is totally bounded, then every sequence in $X$ has a Cauchy subsequence. 

 Sketch of the proof: Let $(x_n)$ be any sequence. By total boundedness, we can cover $X$ with only finitely many balls of radius 1. Since $(x_n)$ has infinitely many elements, there is some ball $B_1$ of radius 1 containing infinitely many elements of the sequence. Let $k_1$ the smallest index such that $x_{k_1} \in B_1$. Notice that $B_1$ is totally bounded as well, so we repeat the argument and find an open ball $B_2 \subset B_1$ of radius $1/2$ which contains infinitely many elements of our sequence, and we let $k_2 > k_1$ be the smallest integer greater that $k_1$ for which $x_{k_2} \in B_2$. In the same way, we find a ball $B_3 \subset B_2$ of radius $1/3$ containing infinitely many elements of the sequence, and an index $k_3 > k_2$ such that $x_{k_3} \in B_3$, and so on. Doing this inductively and appealing to lemma 1 above, we find that the subsequence $(x_{k_n})$ is Cauchy. 

An immidiate corollary is the following:
Theorem 4. If $X$ is totally bounded and complete, then $X$ is sequentially compact. 
So now only one implication is left, and we will prove it by by first looking at another lemma. 
Lemma 3. If $X$ is not compact, then there is some countable open cover $\{U_1, U_2, \ldots\}$ of $X$ which has no finite subcover. 
We can now prove the missing link:
Theorem 5. If $X$ is not compact, it is not sequentially compact. 

 Let $\mathcal{U} = \{U_1, U_2, \ldots\}$ be a countable cover of $X$ without finite subcover. In particular, for any $n$ there is some $x_n$ not contained in $U_1 \cup \cdots \cup U_n$. We claim that $(x_n)$ has no convergent subsequence. To see this, let $x \in X$ be arbitrary. Since $\mathcal{U}$ is a cover of $X$, there is some integer $k$ with $x \in U_k$. By construction, $U_k$ is an open neighbourhood of $x$ which contains at most finitely many points of $(x_n)$. Therefore, $x$ can not be the limit of any subsequence of $(x_n)$, and since $x$ was arbitrary, we see that $(x_n)$ has no convergent subsequence. 

Theorems 1,2, 4, and 5 together imply that $X$ is compact iff it is sequentially comact iff it is totally bounded and complete. The only statements left in your question are those concerning $\mathbb{R}$, and with the theorems proved so far, we are done once we prove the following easy theorems:
Theorem 6 A set $X \subseteq \mathbb{R}$ is totally bounded if and only if it is bounded. 

 Clearly totally bounded implies bounded. For the converse, suppose $X \subset [a, b]$ for certain $a, b \in \mathbb{R}$. Then for any $\epsilon>0$ the finitely many balls $X \cap (a + k\epsilon - \epsilon/2, a + k\epsilon - \epsilon/2)$ with $0 \leq k \leq (b-a)/\epsilon$ cover $X$. 

Theorem 7 A set $X \subseteq \mathbb{R}$ is complete if and only if it is closed in $\mathbb{R}$. 

 Since $\mathbb{R}$ is complete, any Cauchy sequence in $X$ has a limit in $\mathbb{R}$. If $X$ is also closed, then that limit belongs to $X$, so any Cauchy sequence in $X$ converges in $X$. Conversely, if $X$ is complete, then in particular any sequence in $X$ converging in $\mathbb{R}$ converges in $X$ as well, so $X$ contains its boundary, so $X$ is closed. 

A: Here is another viewpoint on your question.  The other poster has given you a quick course on the subject that will reward careful study, so here are some more relaxed reflections on it. 
To a large extent this is a question for textbook writers since that is where you are likely to address such concerns.  In a very advanced textbook that, nonetheless, feels obliged to prove the standard elementary results in metric spaces there might be a question as to which order of presentation is the most economical.  For most of us though this is a pedagogical question and we just try to present an order that makes the concepts compelling.
In our text Elementary Real Analysis (Thomson*Bruckner${}^2$),
in Chapter 13 on metric spaces,
 we do compactness after completeness and after many examples.  Compactness is defined by the Bolzano-Weierstrass property which is the easiest way into the ideas.  There is a whole section exploring the concepts, and a further section studying continuous functions in relation to compactness.  It is only then that we prove that the Heine-Borel property is equivalent to compactness thus defined.  After that there is a section on total boundedness and the characterization of compactness using that idea. We considered these latter two sections as rather more advanced.
For metric spaces this seems reasonable.  In a general topological space the Bolzano-Weierstrass property does not capture the notion of compactness that one wants and so the definition uses the Heine-Borel property.  For that reason many writers and instructors will always use the word "compact" to be defined by the Heine-Borel property and refer to the Bolzano-Weierstrass property as "sequential compactness."  This distinction is messy and unfortunate for metric spaces, but it helps later on when you have the broader topological view.
What about returning to the real line?  On the real line I'm not sure it is useful (even though it is common) to call closed, bounded sets "compact."  On the real line these properties are theorems, not definitions.  How best to prove them?
In our Chapter 4 we do compactness arguments on the real line.  In addition to Heine-Borel and Bolzano-Weierstrass we include the Cantor Intersection Property and the Cousin Covering Argument.  All are equivalent and one can deduce any one of them from the others.
Should we arrange the four properties in one theorem and find the most efficient way to prove the equivalence?  Not at all for a textbook.  What you really want to learn here is how to use any one of these compactness arguments to prove other theorems.  So it is not a question of efficiency.  Instead start with any one of them and prove any other of them.  For any theorem that you can prove with one of them, try proving using a different compactness argument and comment on which you prefer.
By the way, if all that one wants to do is prove Heine-Borel and Bolzano-Weierstrass etc. in the most efficient manner I would do this:


*

*Bounded monotone sequences converge.

*All sequences have monotone subsequences.

*All bounded sequences have monotone convergent subsequences (i.e., BW).

*A nested shrinking sequence of intervals contains a unique point.

*Cousin's lemma follows from this.

*HB follows from Cousins lemma.

*Cantor Intersection Property follows from Cousins lemma.

