This question is old...but I'm still going to give it a go.
Before understanding what a topology is, it is important to understand what a set is without a topology. Without a topology, a set is akin to a sealed bag full of elements: We are on the outside of the bag, and so far as we can tell, each object in the bag is indistinguishable from each other object in the bag; it is easy to see that two, or three, or four objects are unique, but beyond this, it is difficult to truly say anything about any given object in particular. The objects are simply there, and the only property that we can very truthfully assign to the bag (set) itself is the number of objects that the bag (set) contains. In other words, Cardinality is the central, and indeed only, notion which defines a set (in so far as the elements relate to one another).
Of course, in practice, we rarely work with sets whose only property is cardinality. We work with the real line, in which there is a well-defined notion of distance between elements; there is even an order which is imposed on the elements of the set. We work in the Euclidean plane, where there is no longer a well-defined, useful order between arbitrary elements, but there is now a notion of vectors, which have a notion of length attached to them (distance from the origin point), and even of angle between them. The crucial thing to notice here is that all of these very spacial, geometric properties, the added structure which makes sets like Euclidean space so very interesting, are inherently relationships between elements. The set no longer has only cardinality, but now has well-defined ways in which elements relate to one another.
Let us consider how some of these properties relate to one another in the Euclidean plane. The geometry of the Euclidean plane, in many ways, derives from its inner product. From the inner product, we can derive a formula for the cosine of an angle between vectors, and from this comes the notion of angle. We can also define a formula for a norm, a length, of vectors in the space. Thus from angle, we derive the notion of length. From the norm, we are able to derive a formula for distance: We see that a so-called inner product space implies also the structure of a so-called normed linear space, and from this, we find the structure also of a metric space. We are drilling down through increasingly lower-level geometric properties: Angle is a more strongly geometric notion than length, and length is a stronger notion than distance.
This begs the question: What geometry does a space retain when we cannot even measure a notion of distance between elements? What relation is left? What is the minimal geometric relation that a set can be endowed with? Topology offers the answer of nearness. When defining a topology on a set, a mathematician provides explicit neighborhoods for every point: He or she explicitly defines what sets of objects are considered to be near to one another. Each open set in a topology represents this extremely low-level geometric idea, which relies not even on "distance". The amazing thing about point-set topology is that, through its theorems, we come to learn how the way that points are considered near to each other, as well as how they can be distinguished, how they can be separated, and how many of them there are, affect our ability to define such relations as "distance", "angle", "completeness", and "length", which correspond to our intuition for such ideas. Thus, from a geometric perspective, point-set topology largely begins with the explicit declaration of which objects of a set are "near" to each other, and explores the implications that this hierarchy of neighborhoods has on the geometric structure of the set.
Best Answer
$f:[0, 1]\to X$ defined by $$f(x)=\begin{cases} 1\quad \quad \text{$x\in (0,\frac{1}{6})$} &\\ 2\quad \quad \text{$x\in (\frac{1}{5},\frac{1}{4})$}&\\ 3\quad \quad \text{$x\in (\frac{1}{3},\frac{1}{2})$}&\\ 4\quad \quad \text{otherwise} \end{cases}$$
Then $f$ is continuous map as pre-image of every open set is open.
$f^{-1}(\{2\}) =(\frac{1}{5},\frac{1}{4})$
$f^{-1}(\{1,2\}) =(0,\frac{1}{6})\cup (\frac{1}{5},\frac{1}{4}) $
$f^{-1}(\{2,3\}) = (\frac{1}{5},\frac{1}{4})\cup (\frac{1}{3},\frac{1}{2}) $
$\begin{align} f^{-1}(\{1,2,3\}) &= (0,\frac{1}{6})\cup(\frac{1}{5},\frac{1}{4})\cup (\frac{1}{3},\frac{1}{2}) \end{align}$
$f^{-1}(X) =[0, 1]$
$f^{-1}(\emptyset ) =\emptyset$