How to show $f(\Bbb R,\tau_e)\to(X,\tau)$ is continuous? 
Let $\tau =\{\emptyset,X,\{a\},\{c\},\{a,c\}\}$ on  $X=\{a,b,c\}$
and $\tau_e$ is usual topology on $\Bbb R$
$$f(x)=\begin{cases}a,\;\;x<0\\b,\;\;0\leq
 x\leq1\\c,\;\;1<x\end{cases}$$
Show that $f(\Bbb R,\tau_e)\to(X,\tau)$ is continuous

f is continuous iff for every $H\in \tau $ , $f^{-1}(H)\in \tau_e$
Do we need to show  inverse of the every element in X?
I mean for $\{a,c\} \subset \tau $ and $f^{-1}(\{a,b\})=(-\infty,0)\cup(1,\infty)$ which is in $\tau_e$ because it is the union of $(-\infty,0)\in\tau_e$ and $(1,\infty)\in\tau_e$ ?
 A: Yes, you've got the right idea. I think you mean to say that $\{a, c\} \in \tau$, rather than that it's an element of $\tau$ (and you don't want to compute $f^{-1}(\{a, b\})$ either); then we know 
$$f^{-1} \Big(\{a, c\}\Big) = (-\infty, 0) \cup (1, \infty)$$
which is an open set of real numbers. Hence $f^{-1} \Big(\{a, c\}\Big) \in \tau_{\epsilon}$, as desired.
Then you are correct about how to proceed: You need to show that the preimage of each of the $5$ sets in $\tau$ is open; two of these are trivial (as always), and you've already done one of the other three.
A: Yes, you need to show that $f^{–1}(U)\in \tau_e$ for each $U\in\tau$. But this is easy since there are only five such sets, and you get two of them for free.
You've pointed out that two others--$f^{-1}(\{a\})$ and $f^{-1}(\{c\})$--are good, and that the final one (which is their union) is also good, so you are done.
Edit: Note, however, that it is not enough to only show that $f^{-1}(\{a,c\})$ is open. You really do need to consider $f^{-1}(\{a\})$ and $f^{-1}(\{c\})$ separately as well.
