(1) Let $U_0$ and $V_0$ be open sets in $X$ such that $U = U_0 \setminus A$ and $V = V_0 \setminus A$, and suppose that $A \cup U$ is not connected; then there are non-empty sets $H,G \subseteq A \cup U$ such that $A \cup U = G \cup H$, $G \cap H = \emptyset$, and $H$ and $G$ are closed and open in the subspace topology on $A \cup U$.
Suppose that $G \cap A \ne \emptyset \ne H \cap A$. Since $G$ and $H$ are open in $A \cup U$, there are open sets $W_G,W_H$ in $X$ such that $G = W_G \cap (A \cup U)$ and $H = W_H \cap (A \cup U)$. Then $G \cap A $ = $W_G \cap (A \cup U) \cap A = W_G \cap A$, since $U \cap A = \emptyset$, and similarly $H \cap A = W_H \cap A$. Thus, $G \cap A$ and $H \cap A$ are relatively open, non-empty subsets of $A$, and since $A \subseteq G \cup H$, their union is $A$; this contradicts the assumption that $A$ is connected. We may therefore assume that $A \subseteq G$, say, so that $H \cap A = \emptyset$, $W_H \cap A = \emptyset$, and hence $H = W_H \cap U$.
Now $W_H \cap U \subseteq W_H \cap U_0 \subseteq W_H \cap (U \cup A) = (W_H \cap U) \cup (W_H \cap A) = W_H \cap U$, so $W_H \cap U = W_H \cap U_0$, and hence $H = W_H \cap U$ is open in $X$. I claim that $H$ is also closed in $X$, contradicting the assumption that $X$ is connected.
Clearly $X \setminus H = G \cup V = \left(W_G \cap (A \cup U) \right) \cup V = \left(W_G \cap (X \setminus V) \right) \cup V = W_G \cup V$. But $A \subseteq G \subseteq W_G$, so $V_0 \setminus V \subseteq A \subseteq W_G$, and $X \setminus H = W_G \cup V = W_G \cup V_0$, an open subset of $X$, as claimed.
(2) Let $K$ be an uncountable subet of the Sorgenfrey line. Suppose that $x \in K$ and there is a sequence $\langle x_n:n \in \omega \rangle$ in $K$ such that $x_n < x_{n+1}$ for all $n \in \omega$ and $x = \sup\{x_n:n \in \omega \}$ (i.e., the sequence converges monotonically up to $x$ in the Euclidean topology). Let $V_L = (\leftarrow,x_0)$, $V_R = [x, \to)$, and for $n \in \omega$ let $V_n = [x_n,x_{n+1})$; then $\mathcal V = \{V_L,V_R\} \cup \{V_n:n \in \omega\}$ is an open cover of $K$ with no finite subcover. Thus, it suffices to show that $K$ must contain such a point $x$ and sequence $\langle x_n:n \in \omega \rangle$.
If not, for each $x \in K$ there must be a rational number $r(x) such that $(r(x),x) \cap K = \emptyset$. But $\mathbb Q$ is countable, so there must be distinct $x,y \in K$ such that $r(x) = r(y)$. (In fact the function $r$ must be constant on an uncountable subset of $K$, but we need only two points with the same $r$ value.) Without loss of generality $x. But then $r(y) = r(x) < x < y$, and $x \in (r(y),y) \cap K$, contradicting the choice of $r(y)$. Thus, the desired point and sequence must exist, and $K$ cannot be compact (or even countably compact, since $\mathcal V$ is countable).