Nested Sequence of Topological Space and the Coherent Topology; Munkres Topology 35.9 This is problem 9 from chapter 35 in Munkres:

Here's my attempted solution:
a. 
The fact that the coherent set is in fact a topology on $X$ is simple. I won't show it here.
In order to show that $X_i$ is a subspace of $X$ we must show that $\tau_i = \{X_i \cap U | U$ open in $X \}$, where $\tau_i$ is the given topology of $X_i$.
So let $U_i \in \tau_i$. Let $j < i$. Since $X_j$ is a subspace of $X_i$ (by induction), $X_j \cap U_i$ is open in $X_j$. For $i < j$ we have that, again by induction, that $X_i$ is closed in $X_j$. Since $X_i - U_i$ is closed in $X_i$ we have that $X_j \cap (X_i - U_i) = X_j - U_i$ is closed in $X_j$. So $U_i$ is open in $X_j$. It follows that $U_i = X_i \cap U_i$ where $U_i$ is open in $X$. The other direction is direct from the definition.
b. 
Let $f: X \to Y$ be with $f_{|X_i}$ continuous for each $i$.
Take $x \in X$, and let $f(x) \in V$ open in $X$. $\exists x \in X_i$ for some $i$ and $f_{|X_i}(x) \in V$ as well. Since $f_{|X_i}$ is continuous $\exists x \in U_i$ open in $X_i$ s.t $f_{|X_i}(U_i) \subset V$. But $f(U_i) = f_{|X_i}(U_i)$. We've showed above that $U_i$ is open in $X$, so $f$ is continuous. 
c.
Let  $A,B$ be closed and disjoint in $X$. Define $f_0: A \cup B \to [0,1]$ by $f(A) = \{0\}$, $f(B) = \{1\}$ which is continuous. We note that $A \cup B$ is closed in $A \cup B \cup X_1$ because $(A \cup B \cup X_1) \cap (A \cup B) = A \cup B$ and $A \cup B$ is closed in $X$.
By Tietze extend $f_0$ to $f_1:A \cup B \cup X_1 \to [0,1] $. Do so inductively (the above argument works for any $i$), to get $f_i$ a continuous extension of $f_{i-1}$. Define $f: X \to [0,1]$ by $f(x) = f_i(x)$ whence $x \in X_i$, it is well defined since the sequence is of extensions and the spaces are nested. By clause b. $f$ is continuous. We see that $A \subset f^{-1}([0,\frac{1}{2}))$ and that $B \subset f^{-1}((\frac{1}{2}, 1])$, disjoint open neighbourhoods of $X$.
 A: Suppose $C$ is closed in $X$, this means that $U= X\setminus C$ is open in $X$.
This means that for all $i$, $X_i \cap U$ is open in $X_i$ or $X_i \setminus (X_i \cap U) = (C \cap X_i)$ is closed in $X_i$
So the closed sets of $X$ have an analogous description as the open sets of $X$. As the $X_i$ are closed, this is convenient: in all spaces $X$: if $A$ is closed in $B$ and $B$ is closed in $X$, then $A$ is closed in $X$.
$C \subseteq X_i$ is closed in $X_i$, then $C$ is closed in $X_{i+1}$ by the above fact and $X_i$ being closed in $X_{i+1}$ for all $i$. By induction we show that $C$ is closed in $X_j$ for all $j > i$. And by definition $C \cap X_m$ is closed in $X_m$ for all $m < i$ as well. It follows that $C$ is closed in $X$. And the reverse holds by the first paragraph. So the closed sets in $X_i$ are just the closed sets in $X$ intersected with $X_i$. This shows a). 
In your proof, for $U_i \in \tau_i$ you claim that $U_i$ is open in all $X_j$; this does not always hold (e.g. $(0,1]$ is open in $[0,1]$ but not in $[0,2]$ nor $[0,n]$). This is the reason I went for closed sets, as there we do have "closed in closed is closed".
As to b): If $O \subseteq Y$ is open $f^{-1}[O] \cap X_i = f|_{X_i}^{-1}[O]$ is open in $X_i$ as $f|_{X_i}: X_i \to Y$ is continuous. So $f^{-1}[O]$ is open in $X$.  
In your proof $U_i$ need not be open in $X$ (being relatively open in a closed subset, e.g. $(0,1]$ is open in $[0,1]$ but not in the reals).
c) looks OK, though the hint is I think not needed;  choose $f_0:X_0 \to [0,1]$ such that $f_0[A \cap X_0] = \{0\}$ and $f_0[B \cap X_0] = \{1\}$. Then given $f_i: X_i \to [0,1]$ continuous with $f_i[A \cap X_i] = \{0\}$ and $f_i[A \cap X_i] = \{1\}$, and such that $f_i|_{X_{i-1}} = f_{i-1}$ for $i \ge 1$, then define $g : (A \cap X_{i+1}) \cup (B \cap X_{i+1}) \cup X_i$ by setting it to $0$ on the first set, $1$ on the second and $f_i$ on the third. This is a consistent definition, and so the pasting lemma for finitely many closed sets (as all these sets are closed in $X_{i+1}$) says that $g$ is continuous on its (closed) domain, and then its Tietze-extension $f_{i+1}: X_{i+1} \to [0,1]$ is as required for $i+1$. This concludes the recursive construction of the $f_i$. Then $f(x) = f_i(x)$ for $x \in X_i$ is well-defined (as every $f_i$ restricted to a lower-indexed $X_j$ is $f_j$ again, so all is consistent.) and continuous by b) and clearly $f[A]  =\{0\}$ and $f[B] = \{1\}$. Then finish as in your proof. 
As to your proof: why is $A \cup B \cup X_i$ normal, so that Tietze applies? $A, B$ possibly live in all $X_i$? All we know is that $X_i$ is normal.. You just seem to copy the hint without critique or argument. 
