Prove that if $(C_n)$ is a sequence of connected subsets of $X$ such that $C_n\cap C_{n+1}\neq\emptyset$ then $\bigcup C_n$ is connected. Suppose that $(C_n)$ is a sequence of connected subsets of $X$ such that $C_n\cap C_{n+1}\neq\emptyset$ for each $n\in\mathbb{N}.$ It is required to prove that $\bigcup C_n$ is connected. The following is my attempt.
Let $P(n)$ be the statement $\bigcup_{k=1}^n C_k$ is connected. Obviously $P(1)$ is true. Now let $n\in\mathbb{N}$. Suppose $P(n)$. Suppose the function $f:\bigcup_{k=1}^{n+1}C_k\to\{0,1\}$ is continuous where the set $\{0,1\}$ is endowed with its discrete topology. Since $A=\bigcup_{k=1}^n C_k$ is connected and $f\restriction_A$ is continous we have $f\restriction_A$ is constant; say $f\restriction_A=1$. Now let $x\in C_{n+1}$ and $a\in C_n\cap C_{n+1}$. Then $f(x)=f(a)$ because $C_{n+1}$ is connected. But $f(a)=1$ as $a\in A$. Thus $f\restriction_{C_{n+1}}\equiv 1$. Therefore $f$ is constant on $\bigcup_{k=1}^{n+1}C_k$. Hence $\bigcup_{k=1}^{n+1}C_k$ is connected. Now by induction $P(n)$ is true for all $n\in\mathbb{N}$, i.e. $\bigcup_{k=1}^n C_k$ is connected for all $n\in\mathbb{N}$.
Now suppose $\bigcup C_n$ is not connected. Then there exists a continuous surjective function $f:\bigcup C_n\to\{0,1\}$. Thus there exist $a,b\in\bigcup C_n$ such that $f(a)=0$ and $f(b)=1$. So there exist $n_a,n_b\in\mathbb{N}$ such that $a\in C_{n_a}$ and $b\in C_{n_b}$. WLOG suppose $n_a\leq n_b$. Then $a,b\in \bigcup_{k=1}^{n_b} C_k$ and $\bigcup_{k=1}^{n_b} C_k$ is connected and therefore $f\restriction_{\bigcup_{k=1}^{n_b} C_k}$ is constant as it is continuous. Therefore $0=f(a)= f(b)=1$; contradiction. Hence $\bigcup C_n$ is connected. 
Is the above proof alright? Thanks.
 A: The first part with induction is fine. I wouldn't do it differently.
The last part can be done differently as well: define $D_n=\cup_{i=1}^n C_i$. All $D_n$ are connected by the first part, and they all intersect in the non-empty $D_1 = C_1$, so their union $\cup_n D_n = \cup_n C_n$ is also connected..
This way both parts depend on the same "union of intersecting connected sets is connected" fact, which I think is cleaner.
A: Looks good to me. Note that as an alternative to the second paragraph, you could instead invoke the fact that a union of connected subsets that intersect nontrivially is connected (if you've already seen/proven this). That is, if $(U_\alpha)$ is a collection of connected subsets and $\bigcap_\alpha U_\alpha \neq \emptyset $, then $\bigcup_\alpha U_\alpha$ is connected.
A: In your proof, since you have already assumed that $f:\cup_{1}^{n+1}C_k\rightarrow\{0,1\}$ is a continuous function (we know that there exists atleast one such continuous function, hence the assumption) , its clear that $\cup_1^nC_k$ and $C_{n+1}$ being connected sets will map to an individual element each in $\{0,1\}$. Then, if further you assume that they map to distinct elements, like $\cup_1^nC_k$ mapping to $0$ and $C_{n+1}$ mapping to $1$, then contradiction shall arise because $X=(\cup_1^nC_k)\cap C_{n+1}\neq\phi$ and for any $x\in X$, the value of $f(x)$ will become uncertain. Thus, by contradiction, we have the entire set $\cup_1^{n+1}C_k$ mapped to either $0$ or $1$ in $\{0,1\}$. Now, $f(x)$ by virtue of being a continuous function shall preserve connectedness, hence, $\cup_1^{n+1}C_k$ is connected and by your induction argument, $\cup^nC_n$ is connected $\forall$ $n$. Thus I feel that the first paragraph of the proof is well and good but the second one is redundant.
