(a) The pasting lemma applied several times. Or, we can just say that for a closed $B\subset Y$ , $f^{-1}(B)=\cup_{\alpha}(f|A_{\alpha})^{-1}(B)$ (this follows from the fact that $\cup_{\alpha}A_{\alpha}=X$ ) is a finite union of closed sets (this follows from the fact that every $A_{\alpha}$ is closed).

(b) $f(x)=0$ on $[1/(n+1),1/n]$ , $f(x)=1$ on $(-\infty,0]$ are continuous, but $f$ on $(-\infty,1]$ is not. The reason is that if the collection is not finite, the union of preimages as in (a) may be not closed, as is the case for $f^{-1}(\{0\})$ , for example.

(c) Let $B$ be a closed subset of $Y$ . Then $A=f^{-1}(B)=\cup_{\alpha}(f|A_{\alpha})^{-1}(B)$ (this follows from the fact that $\cup_{\alpha}A_{\alpha}=X$ ). Suppose $x\notin A$ . There is a neighborhood $U$ of $x$ such that it intersects only a finite number of sets in the collection: $A_{1},...,A_{n}$ . For each $i=1,...,n$ : $(f|A_{i})^{-1}(B)=S_{i}$ is closed in $A_{i}$ and, therefore, closed in $X$ (as $A_{i}$ is closed). Moreover, $x\notin S_{i}$ . Hence, there is a neighborhood $U_{i}$ of $x$ such that $U_{i}\cap S_{i}=\emptyset$ . The intersection $U'=U\cap\cap_{i}U_{i}$ is a neighborhood of $x$ such that $U'\cap A=\emptyset$ . We conclude that $A$ is closed.