There is totally classic result about the structure of non discrete locally compact topological (non-necessarily commutative) fields $K$, whose proof uses the existence of the Haar measure on the underlying additive group of $K$. The theorem (see Bourbaki, Commutative algebra, chapter VI, paragraph 9, section 4, thorem 1 for instance) is the following :
Theorem. Let $K$ be a locally compact non discrete (not necessarily commutative) field. Note $|\cdot|$ the modulus of $K$ (defined thx to the Haar measure), which is an absolute value on $K$.
If $K$ is of characteristic $0$ and if $|\cdot|$ is not non-archimedean, then $K$ is isomorphic ot $\mathbf{R}$, $\mathbf{C}$ or $\mathbf{H}$.
If $K$ is of characteristic $0$ and if $|\cdot|$ is non-archimedean, then $K$ is an algebra of finite rank of the field $\mathbf{Q}_p$ for some prime number $p$.
If $K$ is of charateristic $p > 0$ ($p$ is prime) it is isomorphic to a field whose center is a field $k((T))$ where $k$ is finite extension of $\mathbf{F}_p$, and $K$ is of finite rank over its center.
The proof of the existence of a (left or right) Haar measure on a locally compact topological group is not that hard, but I would nevertheless like to know if it is possible to derive the structure of non discrete locally compact topological (non-necessarily commutative) fields without using the Haar measure.
Remark 1. All compact spaces are supposed to be Hausdorff.
Remark 2. Bourbaki defines a discrete field as we all can imagine (in Topologie générale, chapter 3, p. 55 examples, 2) as a topological field whose topology is the discrete one, and I guess that in the aforementioned theorem the "non discrete" means "whose topology is not the discrete one". Then in the case 2 with simply $K = \mathbf{Q}_p$ of case $3$ with simply $K = \mathbf{F}_p ((T))$, $K$ may indeed by a totally disconnected topological space, but it is not a discrete topological space.
