Is the subset 1/2^n compact? Is the subset 
$$[{\frac{1}{2^{n}}|n\epsilon \mathbb{N}}] \cup [2]$$ compact
My thoughts: Use Heine-Borel which states any closed and bounded subset is compact. The sequence $$(\frac{1}{2^{n}})$$ is bounded and is closed since the sequence $$(\frac{1}{2^{n}})$$ which is a subset of A, converges to 1∈A. So the subset is compact. Is this correct? 
(Apologies, as I am not great at the coding on this)
 A: A subset is closed if and only if all of its limit points belong to the set. It remains to prove that there are indeed no limit points of this set which lie outside of the set (which isn't terribly hard to do).
Boundedness is trivial here, and as you pointed out, closedness and boundedness together imply compactness in $\mathbb{R}$ with the standard topology.
A: Alternatively, you can use the formal definition of compactness.  If you have an open covering of the set, then some open set covers a neighborhood of $2$ and therefore it must cover all but a finite number of other members of the set.
A: Define the $k^{\text{th}}$ partial sum
$\tag 1 {\displaystyle s_{k}=\sum _{n=0}^{k} 2^{-n}}$
Define
$\tag 2 L = \{ s_k \mid k \in \Bbb N\} \,\bigcup \,\{2\}$
It is not difficult to show that $L$ is contained in an open ball of radius $2$, so $L$ is bounded.
Define $R_{-1} = (-\infty, 1)$ and $R_{0} = (2, +\infty)$.
For integer $n \ge 1$ define
$\tag 3 U_n = \big(s_{n-1},s_{n}\big)$
The union of this family of open intervals.
$\tag 4 U = \displaystyle{\bigcup_{n \ge -1} U_n}$
is open. So the complement of $U$, call it $K$, must be a closed set.
To show that $L = K$, we'll need the following algebraic identity,
$\tag 5 \text{For every integer } k, \quad 2^{-k} - 2^{-(k+1)}  = 2^{-(k+1)}$
Using $\text{(5)}$ we can show, using induction, that every partial sum $s_k$ satisfies
$\tag 6 s_k = 2 - 2^{-k}$
$ L \subset K$:
We have $2 \in L$ and using $\text{(6)}$ that $2 \notin U$; therefore $2 \in K$.
Let $x \ne 2$ belong to $L$. Then $x = s_k$ for some $k$. Since $1 \le s_k \le 2$ and the sequnece $s_n$ is increasing, we can write $s_k \notin U$ and also $s_k \in K$.
$ K \subset L$:
Again, we only have to demonstrate that if $x \in K$ and $x \ne 2$ then $x = s_k$ for some $k$. But since $x \notin U$ we can write $x \in [1,2)$.  Knowing that the real numbers satisfies the axiom of Archimedes we can argue the existence of an integer $k \ge 0$ such that $x \in [s_k,s_{k+1})$. Since $[s_k,s_{k+1}) = U_{k+1} \cup \{s_k\}$ and by assumption $x \notin U$, we must have $x = s_k$.
