Interval $[0,1]$ is neither compact nor connected in the Sorgenfrey line. 
Let $A=[0,1]$. Show that $A$ is neither compact nor connected in the Sorgenfrey line, $\tau_{[,)}$, and that there is no neighborhood of $0$ compact.

For the connectedness part, I thought that $[0,1)$ might work. It is open in $\tau_{[,)}$, and 
$$\mathbb{R}\setminus [0,1)=(-\infty, 0)\cup [1,\infty)= \bigcup_{a\in \mathbb{R}^+} [-a,0) \cup \bigcup_{b\in \mathbb{R}^+}[1,b)$$
which is a union of open sets and therefore $[0,1)$ is also closed. Since there exists a non trivial subset which is both open and closed, then $(\mathbb{R},\tau_{[,)})$ is not connected, and since $[0,1)\subset A$, then A is not connected neither.
For the compactness part, I'm still struggling with the open cover/ finite subcover definition, so I propose as an attempt the following open cover:
$$\mathfrak{A}=\left\{ [ 0,1+1/n)  \mid n\in \mathbb{Z}^+ \right\}$$
which I believe it doesn't have a finite subcover, but I lack of a formal proof of this fact.
The last paragraph would also provide a proof that there is not a compact neighborhood of $0$. Let $[a,b)$ be such a neighborhood, then
$$\left\{ [a,b+1/n) \mid n\in \mathbb{Z}^+ \right\}$$
would also be an open cover of $[a,b)$ that doesn't have a finite subcover, in contradiction with the hypothesis that $[a,b)$ is compact.
Is this proof correct? Thanks in advance!
 A: Your proof that $[0,1]$ is not connected in the Sorgenfrey topology is fine; your argument that it is not compact, however, is not correct. The open cover $\mathfrak{A}$ has the finite subcover $\{[0,2]\}$; indeed, any single member of $\mathfrak{A}$ covers $[0,1]$. However, the open cover
$$\left\{\left[0,1-\frac1n\right):n\ge 2\right\}\cup\{[1,2)\}$$
works: any subcover of it must include the set $[1,2)$, since that’s the only one containing $1$, and it must contain enough of the intervals $\left[0,1-\frac1n\right)$ to cover $[0,1)$. Clearly, however, no finite collection of these intervals is enough, since the union of any finite collection of them is equal to the largest interval in that finite collection.
To prove that $0$ has no compact nbhd, let $U$ be any nbhd of $0$. Then there is an $a>0$ such that $[0,a)\subseteq U$. Use the idea above to find an open cover of $[0,a)$ with no finite subcover, and add to it the set $U\setminus[0,a)$ to get an open cover of $U$.
A: For the non-compactness you can see that in the following manner. $[0,1]$ is homeomorphic with $[-1,1]$. If  $[-1,1]$ is compact, then $[-1,1]^2$ with the product topology, is compact. It is a well known fact that the antidiagonal set $F=${$(x,-x)|x $$\in$$ \mathbb{R}$} is closed and the subspace topology of this set is the discrete topology. $F $$\cap$ $[-1,1]^2$ is an uncountable closed subset of $[-1,1]^2$, thus it must be compact. But the subspace topology of this set is the discrete and because it is uncountable it cant be compact. So $[-1,1]$ is not compact and because of the homeomorphism $[0,1]$ is not compact too.
