The way I always thought of topology was as a way to describe what "nearness" and "farness" mean, i.e. to give a sense of "space" or "location" to a set. The issue here is that this only really has meaning for certain topologies. Examples where there is no obvious notion of convergence actually meaning getting close to anything would be the cofinite topology (i.e. $(X, \tau)$, where $\tau$ is the set of all subsets $S \subseteq X$ such that $X \setminus S$ is finite), or topologies like $\tau = \tau_{x} = \{ S \subseteq X : x \in S \}$. Our topology professor told us that unless your space is Hausdorff (i.e. for any two points $x, y \in X$, there exist open sets $U, V \in \tau$ such that $x \in U$, $y \in V$, and $U \cap V = \emptyset$), then you don't really have a topology that conveys these ideas of "space".

When you're not Hausdorff, there are still cool things that can be done with a topology (for example, you can show that the Tychonoff's theorem is equivalent to the axiom of choice). Alternatively, though this may be a stretch, you'll sometimes find that you can use other objects to induce topologies which are not themselves Hausdorff, but you'll find that the induced topology might convey other sorts of information.

One example is to consider a complete probability space $(X, \mathcal{F}, \mu)$, i.e. a probability space such that if $\mu(F) = 0$, and $E \subseteq F$, then $\mu(E) = 0$ (note that a Borel probability space is not generally complete). In this case, if we define $\tau = \{ F \subseteq X : \mu(F) = 1 \} \cup \{ \emptyset \}$, then we have a topology. Now, in general, this will not meet many of your separation axioms. For example, if we consider the probability space $([0, 1], \mathcal{L}, \lambda)$, where $\lambda$ is Lebesgue measure and $\mathcal{L}$ is the completion of the Borel $\sigma$-algebra, then you do not have separability. We can check that it's $\textrm{T}_{1}$, but it will not be $\textrm{T}_{2}$. To see this, assume for contradiction that $x, y \in [0, 1]$, and that $x \in U, y \in V, U \cap V = \emptyset, \mu(U) = \mu(V) = 1$. Then $\mu(U \cup V) = \mu(U) + \mu(V) = 2 > \mu([0, 1])$, a contradiction. That no topology generated this way is $\textrm{T}_{2}$ will follow in the same manner.

Second, let $\mathcal{U}$ be an ultrafilter, that is, a family of subsets of $X$ such that (i) every superset of an element of $\mathcal{U}$ is an element of $\mathcal{U}$, (ii) every finite intersection of elements of $\mathcal{U}$ is in $\mathcal{U}$, and (iii) for every subset $S \subseteq X$, either $S \in \mathcal{U}$, or $S^{\complement} \in \mathcal{U}$, but not both. It is trivial to see that $\tau = \mathcal{U} \cup \{ \emptyset \}$ is a topology (the closure of finite intersections is guaranteed by (ii), and the closure under arbitrary union is promised by (i)). In this case, we can show that such a topology is never Hausdorff, again because we can't have disjoint open sets. Suppose for contradiction that $x \in U \in \mathcal{U}, y \in V \in \tau, U \cap V = \emptyset$. Then since neither $U$ nor $V$ is empty, we have that both must be elements of $\mathcal{U}$. Thus $V \subseteq U^{\complement} \Rightarrow V^{\complement} \supseteq U \Rightarrow V^{\complement} \in \mathcal{U}$. But this means that both $V$ and $V^{\complement}$ are elements of $\mathcal{U}$, contradicting axiom (iii).

But what both topologies convey, i.e. the properties which define the open sets $\tau$, are one of those that are conveyed by, say, $\mathbb{R}$ with its standard metric, and that is a certain sense of *bigness*. A set of probability $1$ is big in the sense of measure, and an element of an ultrafilter is big in the sense of the ultrafilter. Similarly, an open set in $\mathbb{R}$ is "big" in the sense that it surrounds all its points, whereas a non-open set would not. Again, this might be a stretch, but it does suggest that even a non-Hausdorff space can be useful.