Suppose $A,B$ are disjoint closed sets in the Sorgenfrey line. We have to find open sets $U,V$ such that $A\subseteq U,\ B\subseteq V,$ and $U\cap V=\emptyset.$
For each $a\in A$ choose $a'\gt a,$ $\ [a,a')\cap B=\emptyset$; then $U=\bigcup\limits_{a\in A}[a,a')$ is an open set containing $A.$
For each $b\in B$ choose $b'\gt b,$ $\ [b,b')\cap A=\emptyset$; then $V=\bigcup_\limits{b\in B}[b,b')$ is an open set containing $B.$
To show that $U\cap V=\emptyset,$ it suffices to show that $[a,a')\cap[b,b')=\emptyset$ for all $a\in A,\ b\in B.$ Suppose $a\in A,\ b\in B,$ and assume without loss of generality that $a\lt b.$ Since $[a,a')\cap B=\emptyset,$ it follows that $b\ge a'$ and so $[a,a')\cap[b,b')=\emptyset.$
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 $U$.
Best Answer
Let $A$ be a subspace of the Sorgenfrey line $S$ and let $a,b\in A$ with $a<b.$ Then $[b,\infty)$ and $(-\infty,b)=\cup_{x<b}[x,b)$ are open in $S.$ So $A\cap (-\infty,b)$ and $A\cap [b.\infty)$ are relatively open in $A,$ are disjoint, and are not empty, and their union is $ A.$ So $A$ is not connected. So the only connected subspaces of $S$ are the empty set and the $1$-element subsets.
Any open $B\subset S$ such that $(b,c)\subset B$ and $b\not \in B$ is not closed, because any nbhd $U$ of $b$ covers $[b,d)$ for some $d>b,$ so $U\cap B\supset (b,\min (c,d))\ne \emptyset.$ So $b\in \bar B$ \ $B.$ This however is not necessary in order that $B$ be open but not closed. For example if $B=\cup_{n\in N} [\frac {1}{2n},\frac {1}{2n-1})$ then $0\in \bar B$ \ $B.$ An open $B\subset S$ is not closed iff there exists $b\in S$ \ $B$ such that $b=\inf\; ((b,\infty)\cap B\;).$