The complex conjugate is not holomorphic

Counterexample

Let $f : \mathbb{C} \to \mathbb{C}$ be the complex conjugate map, $f(z) = \bar z$. The function $f$ is continuous, but it is not complex differentiable at any point $z_0 \in \mathbb{C}$. Thus $f$ is not holomorphic on any nonempty open set. Proof: the difference quotient at $z_0$ is $$\frac{f(z_0 + h) - f(z_0)}{h} = \frac{\bar h}{h}.$$ It is $1$ for each real $h \neq 0$ and $-1$ for each purely imaginary $h \neq 0$. So it has no limit as $h \to 0$.

Remarks

The real and imaginary parts $u = x$ and $v = -y$ are differentiable, but $u_x = 1$ and $v_y = -1$. So the Cauchy–Riemann equations fail at each point. As a map of $\mathbb{R}^2$, $f$ is the reflection in the real axis. It is real linear but not complex linear. For the same reason, $\operatorname{Re} z$, $\operatorname{Im} z$ and $|z|$ are not holomorphic on any nonempty open set.

Explanations

L1 human

As a map of the plane, conjugation is a mirror, and it is very smooth. But a complex derivative must be the same in each direction, and conjugation fixes the real direction and reverses the imaginary direction. So the quotient gives $1$ in one direction and $-1$ in the other, and there is no complex derivative.