You're not wrong in thinking that the definitions of "topological space" and "measurable space" are similar! You're also not wrong in thinking that these similar axiomatizations might lead to "similar looking" structures, and the question is "can we make this precise?". The answer is, of course, "yes". Analyzing the similarities between constructions is one of the things Category Theory excels at, and indeed we can apply it here. In a way that we're about to make precise, the category of topological spaces (with continuous maps) $\mathsf{Top}$ and the category of measurable spaces (with measurable maps) $\mathsf{Meas}$ are very similar.
The key notion is that of a topologically concrete category. This is a category $\mathcal{T}$ equipped with a faithful functor $U : \mathcal{T} \to \mathsf{Set}$ that has certain nice properties inspired by the nice properties of the "Underlying Set" functor $U : \mathsf{Top} \to \mathsf{Set}$. You can read the all about them at the linked article, or in Adamek, Herrlich, and Strecker's The Joy of Cats, but I'll summarize a few important properties of topological categories here:
They're complete and cocomplete. Moreover, we can construct limits/colimits in $\mathcal{T}$ by looking at the limit/colimit of underlying sets, then choosing "the right topology" (resp. $\sigma$-algebra, etc).
It's possible that multiple spaces in $\mathcal{T}$ will have the same underlying set. We end up with a (possibly large) complete lattice of spaces over any set $X$. That is, given any family of topologies (resp. $\sigma$-algebras, etc) on $X$, there is a unique largest topology (resp. $\sigma$-algebra, etc) contained in each member of that family, and a unique smallest topology (resp. $\sigma$-algebra, etc) containing each member of that family.
As a particular example of (2), every set $X$ has a discrete and indiscrete topology (resp. $\sigma$-algebra, etc). Moreover, every map from a discrete space is continuous (resp. measurable, etc) and every map to an indiscrete space is continuous (resp. measurable, etc). We can phrase this succinctly by saying that $U$ has left and right adjoints $\Delta \dashv U \dashv \nabla$ (sending a set to its discrete or indiscrete space respectively).
Every morphism in $\mathcal{T}$ factors as a surjection (epi) followed by a subspace inclusion (regular mono).
There's much more to say about the nice properties enjoyed by topologically concrete categories, but suffice to say this framework provides a robust way of comparing $\mathsf{Top}$ with $\mathsf{Meas}$ -- categorically they are extremely similar!
I hope this helps ^_^
As Martin Brandenburg comments, a topology in which the intersection of every family of open sets is open is known under the name Alexandrov topology.
By definition each Alexandrov topology is a topology, but the converse is not true. Note, however, that the discrete topology is always an Alexandrov topology. But as you say, this topology is not very interesting.
To each topology $\tau$ one can assign the Alexandrov topology $\tau^*$ generated by $\tau$. This is the coarsest Alexandrov topology containing $\tau$ (i.e. the intersection of all Alexandrov topologies containing $\tau$). Thus $\tau^*$ is always finer than $\tau$.
When is $\tau^*$ the discrete topology?
Fact 1. An Alexandrov topology $\vartheta$ is discrete if and only if the space $(X,\vartheta)$ is a $T_1$-space.
Trivially each discrete space is $T_1$. Conversely, if $(X,\vartheta)$ is $T_1$ (which is equivalent to all one-point subsets being closed), then all subsets are closed because they are unions of one-point subsets. Note that all unions of closed subsets are closed in Alexandrov topologies. This means that all subsets are open, i.e. $\vartheta$ is discrete.
Fact 1 shows that Alexandrov topologies are not very interesting if we require a very mild separation axiom.
Fact 2. If a topological space $(X,\tau)$ is $T_1$ , then $(X,\tau^*)$ is $T_1$ and hence discrete.
This is true because $\tau^*$ is finer than $\tau$.
The two above facts show that it would not be a good idea to adopt closedness under arbitrary intersections of open sets as a general axiom for topological spaces. For example, none of the standard spaces occuring in analysis would be a topological space in that sense.
Best Answer
Topologies and $\sigma$-algebras are designed with different objectives in mind. $\sigma$-algebras are designed to play nicely with measures, which are a generalized kind of volume measuring map. Topologies are designed to capture a notion of "closeness": when is a point $x$ close to a set $S$? If every open neighborhood of $x$ intersects $S$. When does a sequence get arbitrarily close to $x$? If every open neighborhood of $x$ contains points in the sequence. Stuff like that. So it's not surprising that at the outset, topologies and $\sigma$-algebras are different.
But! If we think about it some more, then we might find that intuitively, the open neighborhoods of a point are those which have a certain volume. Like, if I put an open ball around $x$, I can tell that it has a non-zero volume. And $\sigma$-algebras are designed to allow volume measurements. So shouldn't all the open sets somehow be made into a $\sigma$-algebra? After all, it might come in handy to assign a volume to such sets. And the answer is yes, that makes sense. We would like it a lot if we could assign a volume to open sets. For instance, this would allow continuous functions to play nicely with volume, since continuous functions play nicely with open sets. And that's why we define the Borel $\sigma$-algebra: given a topological space $(X,\tau)$, we define the Borel $\sigma$-algebra on $X$ as $\mathcal B(X):=\sigma(\tau)$, that is the smallest $\sigma$-algebra containing all the open subsets of $X$, so all the subsets which should have volume. Now $(X,\mathcal B(X))$ is a measurable space on which we could define a measure $\mu$ to assign a volume to each open set, if we were so inclined. This approach is often taken to define the Lebesgue measure, for instance. We take each open set of $\mathbb R^n$ and assign it the volume it should intuitively have, and then we take all the other sets we might get by uniting and intersecting these and assign them a volume which is in line with the definition of a measure. (There is a "better" approach using outer measures which yields more measurable sets, but this one is simpler.)
But the Borel $\sigma$-algebra is just one specific $\sigma$-algebra we might want. For other applications, different ones might work better, especially if we don't actually care about a sense of closeness on the underlying set. Then we don't need a topology, so why restrict our $\sigma$-algebra with a topology?