Continuity of the derivative

As we all know, all the basic properties of holomorphic functions (i.e. functions which are differentiable in the complex sense) can be deduced from Cauchy's formula. Moreover, Cauchy's formula itself can be viewed as a rather simple consequence of the Green-Riemann formula, provided that the holomorphic function you have at hand is assumed to have a continuous derivative.

Of course, Cauchy's formula holds without assuming continuity of the derivative, and it yields continuity of the derivative and much more since it implies power series expansion. But Cauchy's formula (or, if you prefer, Cauchy's theorem) without assuming continuity of the derivative is a rather subtle thing, and this gives a rather "indirect" proof of the fact that holomorphic functions are in fact $\mathcal C^1$.

So my question is the following: does anybody know a direct proof of the fact that if a function $f$ defined on an open subset of $\mathbb C$ is differentiable in the complex sense, then its derivative $f'$ is continuous?

I'm pretty sure I am not the first one and will not be the last one to ask this question, at least for himself (or herself). So please feel free to close it if it has indeed been asked previously on this site.

Edit. Perhaps I should say a few more words about what I mean by a "direct proof". Anything that relies in one way or another to Cauchy's formula or Cauchy's theorem is not considered as a direct argument. A direct proof should somehow establish "from scratch", or "from very basic principles" that holomorphic implies $\mathcal C^1$.

There is an efficient and straightforward proof of the Cauchy-Goursat theorem, by a `lion-hunting' argument based only on differentiability at every point in $\Omega$, available at this link.

In an honors undergraduate analysis class that has dealt with path integrals, one can use it to provide an account of the Cauchy integral formula, analyticity and the classification of zeros and poles in only a couple of weeks.

Does this link help you? Is is entitled: "A TOPOLOGICAL PROOF OF THE CONTINUITY OF THE DERIVATIVE OF A FUNCTION OF A COMPLEX VARIABLE": https://projecteuclid.org/download/pdf_1/euclid.bams/1183522992

In the demonstration no use is made of complex integration, only results from differential topology and Rouche's theorem to arrive to: If f'(z) exists for all z in an open connected set E of the complex plane, then f'(z) is continuous in E.

This does seem to require the Cauchy-Goursat theorem. And, really, there is some strong intuition in this. If you have an open region $\Omega$ where $f$ is differentiable, then $$ \mu(R)=\oint_{R}f(z)dz $$ defines a finitely-additive set function on solid rectangles $R$ contained in $\Omega$ with the property that $$ \lim_{|S|\rightarrow 0}\frac{\mu(S)}{|S|}=0 $$ for sequences of squares $S$ which contain a common point $z$ (Here $S$ denotes the usual area measure of $S$.) Intuitively, there's just nothing such a complex measure could be but 0 when the derivative of $\mu$ with respect to usual area measure is 0 everywhere. Everything after that is contour integral magic a la Cauchy. I don't think you can get away from the measure-theoretic aspect of this.