Why do we need topological spaces?

The point Preuss makes is that we cannot find a metric $d$ on the set of function such that "$f_n \to f$ pointwise" is equivalent to "$f_n \to f$ in the metric $d$" or $d(f_n,f) \to 0$ etc. Uniform convergence does correspond to a metric (from the supremum-norm). But we can define something more general, a topology, such that we can define $f_n \to f$ in that topology , and moreover in such a way that it exactly corresponds with "$f_n \to f$ pointwise".

The following deficiency of topologies is that a metric defines a topology (but not always conversely) but whereas in metric topologies sequences actually suffice to completely describe that topology, in general topologies this is no longer the case and the familiar (from analysis/calculus) sequence must be replaced by a more general notion of convergence, sequential continuity does no longer suffice (we need general continuity), sequentially compact has to be replaced by general compactness; all of these are mostly improvements (in the sense that the general properties behave better wrt topological constructions), but less familiar (in topology the idea to call a set "compact(like)" iff every sequence has a convergent subsequence, comes from analysis and is older (and often more directly applicable too)).

If you consider the set of all functions $f\colon I\to \mathbb R$ on some interval $I$ (just to fix some context), then you can consider (at least) two different meanings to the convergence $f_n\to f$ of a sequence of functions to a function. One is pointwise convergence, where we say that $f_n\to f$ precisely when $f_n(x)\to f(x)$ for all $x\in I$. Another is uniform convergence which is a much stronger property: for all $\varepsilon >0$ there exists $n$ such that $\sup |f_n(x)-f(x)|< \varepsilon $ for all $x\in I$. Uniform convergence implies pointwise convergence but not vice versa (e.g., $f_n(x)=x^n$ on the interval $I=[0,1]$).

Now, in any metric space, there is a notion of convergence of a sequence. For the set of all functions as above, there is a metric such that convergence according to it is exactly the same as uniform convergence. However, generally, there is no metric that in the same way captures pointwise convergence. Similarly, in a topological space, there is a notion of convergence of a sequence. For the set of functions as above, there is a topology that captures uniform convergence. There is also (another) topology that captures pointwise convergence. In this sense, metric spaces are too stringent to capture all useful notions of convergence. Topological spaces are much more flexible.

It should be noted that the root of the difficulty is in demanding that that the metric function takes values in $\mathbb R$. This is a rather unnatural requirement. A more axiomatic approach would replace $\mathbb R$ by a suitable structure defined by means of properties rather than a particular model. When that is done (e.g., "A note on the metrizability of spaces", Algebra Universalis, 2015) one recovers precisely all topological spaces as such generalized metric values where the metric functions take values in what is known as a (Flagg) value quantale. In that sense, the difference between classical metric spaces and topological spaces is that, in the former, one insists on using the value quantale of real numbers.

At the end of the first chapter of Willard's text "General Topology" (1970), Willard provides the following elegantly stated motivation for the theory ...

$\qquad$The process which topology evolves from, outlined in the next section and the notes, is basic to any pure mathematical discipline. We wish to study a particular property enjoyed by some objects of interest (in this case, continuity of functions on some space) and the efficient way to proceed is to first clean the structure on the space down to the bare bones needed for introducing and developing the property we want. The passage to such abstraction has several well-documented advantages. Among them:

$\qquad$1. Since we have only what is essential, our proofs use only what is essential and thus clarify the nature of the object of study, and the logical dependence of the theorem in question.

$\qquad$2. Proofs become easier. Actually, this is a popular professional myth, with an element of truth. Occasionally, a proof really does get easier as a theorem gets more abstract, but this is offset by the need for more and more interpretive skill on the part of those who would use the theorem. What people really mean when they say "proofs become easier" is something like this: "by establishing some notation and introducing the right definitions and conventions, we can draw together all the theorems about this subject and find common characteristics and even repetitions in their proofs, then prove lemmas which enable us to write large numbers of proofs more succinctly." If the subject matter is carefully chosen, the work done in abstracting the properties needed, establishing notation and proving those lemmas will be more than paid for by the gain in succinctness and clarity of the proofs later on, and by the acquisition of powerful methods for continued investigation of the original objects of study.

$\qquad$Such is the case with topology.

Followup:

As regards sequential convergence . . .

In a metric space, you can define the notion of sequential convergence, and the notion of "open sets", and given two metric spaces $X,Y$ and a function $f:X\to Y$, you can show that the following statements are equivalent:

$(1)\;\,f$ is continuous.

$(2)\;\,$If a sequence $(x_n)$ of elements of $X$ converges to a point $x\in X$, then the sequence $(f(x_n))$ converges to $f(x)$.

$(3)\;\,$If $U$ is an open subset of $Y$, then $f^{-1}(U)$ is an open subset of $X$.

A topological space requires a notion of "open sets" satisfying some specified properties, but the concept of open sets does not depend on the existence of an associated metric.

Given a topological space $X$, we can define sequential convergence as follows:

A sequence $(x_n)$ of elements of $X$ converges to a point $x\in X$ if for each open subset $U$ of $Y$ with $x\in U$, we have $x_n\in U$ for all sufficiently large $n$.

Given two topological spaces $X,Y$ and a function $f:X\to Y$, we define the notion of continuity as follows:

$f$ is continuous if for every open subset $U$ of $Y$, $f^{-1}(U)$ is an open subset of $X$.

With that definition, a function between two metric spaces which was continuous in the metric space context is still continuous in the topological context.

Now suppose we have two topological spaces $X,Y$ and a function $f:X\to Y$.

Consider the statements:

$(1)\;\,f$ is continuous.

$(2)\;\,$If a sequence $(x_n)$ of elements of $X$ converges to a point $x\in X$, then the sequence $(f(x_n))$ converges to $f(x)$.

Then in all cases we have $(1)$ implies $(2)$, but in some cases, the converse can fail.

In other words, sequential convergence is not enough to define the general topological concept of continuity.