§23: write out 1, 2, 3, 4, 5, 8, 10
Suppose $X$ under $\mathcal{T}'$ is connected. Then there exists no pair of disjoint nonempty sets $A, B \in \mathcal{T}'$ that union to give $X$. Then since $\mathcal{T} \subset \mathcal{T}'$, there is also no such pair in $\mathcal{T}$, so that $X$ is connected under $\mathcal{T}$.
Suppose $X$ under $\mathcal{T}$ is connected. Then there is no separation $A, B \in \mathcal{T}$, but we do not know whether there is a separation in $\mathcal{T}'$, hence we cannot say anything about its connectedness.
For example, let $X = \{ a, b \}$, and $\mathcal{T} = \{ \{a \}, \varnothing, X \}$, so that $X$ is connected under $\mathcal{T}$. Then let $\mathcal{T}'$ be equal to $\mathcal{T}$, and let $\mathcal{T}''$ be the discrete topology. Both topologies are supersets of $\mathcal{T}$, but $\mathcal{T}'$ is connected while $\mathcal{T}''$ is not.
Thus if $\mathcal{T}'$ is connected then so is $\mathcal{T}$, but if $\mathcal{T}$ is connected we do not know about $\mathcal{T}'$.
Note that
$$ \bigcup A_n = \bigcup_{n=1}^\infty \left( \bigcup_{i=1}^n A_i \right) $$
By induction, we can show that $\bigcup_{i=1}^n A_i$ is connected for all $n$:
Base case. By assumption, $A_1$ is connected.
Suppose true for $n=k$. Then note that
$$ \bigcup_{i=1}^{k+1} A_i = \left( \bigcup_{i=1}^k A_i \right) \cup A_{k+1} $$
And since $A_k$ and $A_{k+1}$ have a point in common, the two sets above have a point in common, hence the ($k+1$)-th case is true.
Thus every $\bigcup_{i=1}^n A_i$ is connected. But note that $\bigcup A_n$, per the equation above, is just the union of all such sets, which all have the arbitrary point $a \in A_1$ in common. Hence by theorem 23.3, $\bigcup A_n$ is connected.
Note that
$$ A \cup \left( \bigcup_\alpha A_\alpha \right) = \bigcup_\alpha (A \cup A_\alpha) $$
by associativity of the union operation.
Then note that each $A \cup A_\alpha$ is a pair of connected spaces that have a point in common, hence it is open.
Then the right side of the above equation is the union of connected spaces that has the arbitrary point $a \in A$ in common, thus it is connected.
Suppose $X$ is infinite and let $A \cup B$ be a separation of $X$. Thus $A$ and $B$ are disjoint, nonempty, and open. By the finite complement topology, since $A$ is open, $X - A = B$ is finite or is all of $X$. It cannot be all of $X$ because then $A$ would be empty. Thus $B$ is finite. By an analogous argument, $A$ is also finite. Then $X$ is the union of two finite sets, which contradicts the fact that $X$ is infinite.