Show that the set $[0,2] \setminus \{1\}$ is not compact by exhibiting a cover of open intervals which has no finite subcover. I think 
$\bigcup_{k=1}^\infty \left(\frac{-1}{k},2-\frac{1}{k}\right)$
will work, but I'm unsure if the interval includes $2$ as $k \rightarrow \infty$.
 A: Hint: any obstruction to compactness won't be at the 'closed' side. Try creating an open cover that includes more and more elements closer and closer to $1$ with each set, but never quite reaches it.
A: The key is to use the fact that you don't need to include $1$. So you might want to try to utilize intervals of the form 
$$ ( 1 + \frac{1}{n}, 2 + \frac{1}{n}) $$ 
A: Let $K=[0,2] \backslash \{1\}=[0,1[ \cup ]1,2]$. If we put: $I_{k}=(-\frac 1k,1-\frac 1k)$ and $J_k=(]1+\frac 1k,2+\frac 1k)$ then we have :
$$\bigcup_{k=1}^{\infty}(I_k \cup J_k)= K$$
Suppose that: $$K=(\bigcup_{i \in I} I_i ) \cup (\bigcup_{j \in J} J_j)$$
where $I$ and $J$ are finite non empty subsets of $\mathbb N^*=\{1,2,...,\}$
Let $$N=1+\max(I \cup J).$$
Since $x=1-\frac 1N \in K$, we have $x=1-\frac 1N \in I_k $ where $k \in I$. (in fact, $x$ can not be in  a certain $J_j$ because $J_j \subset ]1,2] $ and $x < 1$.)   that mines : $-\frac 1k< 1-\frac 1N < 1-\frac 1k$, then : $N < k$ that is not possible since $N=1+\max(I \cup J)$ and $k \in I$.
