Let $M=\bigcup _{A\in 𝒞}A$ and let $x\in \bigcap _{A\in 𝒞}A\text{.}$
Proof by contradiction.
Assume $M$ is not connected.
Let $B$ and $C$ be open sets such that $$M\subseteq B\cup C,M\cap B\cap C=\varnothing ,M\cap B\ne \varnothing \text{and}M\cap C\ne \varnothing \text{.}$$ Then $x\in B$ or $x\in C\text{.}$
Assume $x\in B\text{.}$
Let $U\in 𝒞$ be such that $C\cap U\ne \varnothing \text{.}$
Since $x\in \bigcap _{A\in 𝒞}A$ then $x\in U,$ so that $$x\in B\cap U\text{and}B\cap U\ne \varnothing \text{.}$$ Since $U\subseteq M$ then $$U\subseteq B\cap C\text{and}U\cap B\cap C=\varnothing \text{.}$$ This is a contradiction to $U$ be connected.
So $M$ is connected.

$\square$

Proof by contradiction.
Assume $\bar{A}$ is not connected.
Let $M$ and $N$ be open subsets of $\bar{A}$ such that $$M\cup N=\bar{A},M\ne \varnothing ,N\ne \varnothing \text{and}M\cap N=\varnothing \text{.}$$ Then $$(M\cap A)\cup (N\cap A)=A\text{and}(M\cap A)\cap (N\cap A)=\varnothing \text{.}$$ There exists $x\in \bar{A}\cap M,$ $x$ is a close point of $A$ and, since $M$ is open, $M$ is a neighbourhood of $x\text{.}$
So $M\cap A\ne \varnothing \text{.}$
There exists $y\in \bar{A}\cap N,$ $y$ is a close point of $A$ and, since $N$ is open, $N$ is a neighbourhood of $y\text{.}$
So $N\cap A\ne \varnothing \text{.}$
This is a contradiction to $A$ is connected.
So $\bar{A}$ is connected.

$\square$

(a) This follows from Example 1.

(b) By Example 2, $\overline{C_x}$ is a connected set that contains $x\text{.}$ So $\overline{C_x}\subseteq C_x\text{.}$ So $\overline{C_x}=C_x\text{.}$

(c) Assume $y\in C_x\text{.}$ Then $C_x$ is a connected set containing $y\text{.}$ So $C_x\subseteq C_y\text{.}$ Then $C_y$ is a connected set containing $x\text{.}$ So $C_y\subseteq C_x\text{.}$ so $C_x=C_y\text{.}$

(d) Assume $x,y\in X$ and $C_x\cap C_y\ne \varnothing \text{.}$ Let $z\in C_x\cap C_y\text{.}$ So $z\in C_x$ and $z\in C_y\text{.}$ By (c), $C_x=C_z=C_y\text{.}$ So $C_x=C_y\text{.}$

$\square$

$\Rightarrow$ To show: If there exists a continuous surjective function $f:X\to \{0,1\}$ then $X$ is not connected.
Assume that $f:X\to \{0,1\}$ is a continuous surjective function.
Let $$A=f^{-1}\left(0\right)\text{and}B=f^{-1}\left(1\right)\text{.}$$ Since $f$ is continuous, $A$ and $B$ are open.
Since $f$ is surjective, $f^{-1}\left(0\right)=A\ne \varnothing$ and $f^{-1}\left(1\right)=B\ne \varnothing \text{.}$
Then $$A\cup B=f^{-1}\left(\{0,1\}\right)=X\text{and}A\cap B=f^{-1}\left(0\right)\cap f^{-1}\left(1\right)=\varnothing \text{.}$$ So $X$ is not connected.

$\Leftarrow$ To show: If $X$ is not connected then there exists a continuous surjective function $f:X\to \{0,1\}\text{.}$
Assume $X$ is not connected.
Then there exist open sets $A$ and $B$ such that $$A\cup B=X,A\ne \varnothing ,B\ne \varnothing \text{and}A\cap B=\varnothing \text{.}$$ Define $f:X\to \{0,1\}$ by $$f\left(x\right)=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}x\in A,\\ 1,& \text{if}\hspace{0.17em}x\in B\text{.}\end{array}$$ Since $X=A\cup B$ and $A\cap B=\varnothing ,$ $f$ is well defined.
Since $A\ne \varnothing$ and $B\ne \varnothing ,$ $f$ is surjective.
Since $A=f^{-1}\left(0\right)$ and $B=f^{-1}\left(1\right)$ are open, $f$ is continuous.
So there exists a continuous surjective function $f:X\to \{0,1\}\text{.}$

$\square$