I would contend that what makes complex analytic (holomorphic) functions so special is the structure of the complex numbers themselves. The fact that the complex numbers are essentially $\mathbb{R}^2$ and are also a field is a small miracle. This miracle is at the heart of the special behavior of complex analytic functions.
It is the two dimensional nature of the plane and the field multiplication allow us to see the C-R equations. My favorite way to see the C-R equations is to consider $$f'(z) =\lim_{\Delta z\to 0} \frac{f(z+\Delta z)-f(z)}{\Delta z} $$ Taking in turn $\Delta z =\Delta x$ and $\Delta z =i \Delta y$ the C-R equations just appear. This requires two dimensions for the two approaches and utilizes that $\mathbb{C}$ is a field when dividing by $\Delta z$. (Other chain rule type proofs of the C-R use both these facts, but are somewhat more subtle about it.)
Surely the Cauchy-Integral Formula, which tells us that a holomorphic function is in fact $C^\infty$ deserves mention, in fact it too is a consequence (less directly) of the two dimensional nature of $\mathbb{C}$ and field structure. The standard proof is to use Cauchy's Integral Theorem, which says that if $f$ is a holomorphic on a simply connected domain, then $$\int_\gamma f =0$$ for any sufficiently nice closed $\gamma$ curves in the domain. This itself is seen by applying Green's (2-d real structure again) and the C-R (field structure).
There are many, many more special and cool behaviors of holomorphic functions, but all of them seem to sit on the miracle of the two structures of the complex numbers. Just to advertise a few, Identity Principle (being 2-d connected is easier than being 1-d connected), zero counting theorems like Rouche's (line integrals sometimes have cool answers), the open mapping theorem (zero counting theorems are super cool), Liouville's Theorem (Cauchy integral theorem again), on and on and on.
Let $g:[\frac{1}{8},\frac{7}{8}]\to \mathbb{R}$ be defined by
$\displaystyle{g(x) = f(x+{\small{\frac{1}{8}}}) - f(x-{\small{\frac{1}{8}}})}$.
The goal is to show that $g(c) = 0$, for some $c$.
If $g(\frac{k}{8})=0$, for some $k \in \{1,3,5,7\}$, then let $c=\frac{k}{8}$, and we're done.
Suppose then that $g(\frac{1}{8}),g(\frac{3}{8}),g(\frac{5}{8}),g(\frac{7}{8})$ are all nonzero.
\begin{align*}
\text{Then}\;&
g({\small{\frac{1}{8}}})+
g({\small{\frac{3}{8}}})+
g({\small{\frac{5}{8}}})+
g({\small{\frac{7}{8}}})
\\[4pt]
=\;\,&
\left(f({\small{\frac{2}{8}}}) - f(0)\right) +
\left(f({\small{\frac{4}{8}}}) - f({\small{\frac{2}{8}}})\right) +
\left(f({\small{\frac{6}{8}}}) - f({\small{\frac{4}{8}}})\right) +
\left(f(1) - f({\small{\frac{6}{8}}})\right)
\\[4pt]
=\;\,&f(1) - f(0)\\[4pt]
=\;\,&0\\[4pt]
\end{align*}
Since $g(\frac{1}{8}),g(\frac{3}{8}),g(\frac{5}{8}),g(\frac{7}{8})$ sum to zero, and since they're all nonzero, it follows that at least one of them is positive, and least one is negative.
Since $f$ is continuous on $[\frac{1}{8},\frac{7}{8}]$, so is $g$.
Hence, by the Intermediate Value Theorem, we must have $g(c) = 0$, for some $c$.
This completes the proof.
Best Answer
The Extreme Value Theorem sits in the middle of a chain of theorems, where each theorem is proved using the last.
LUB Property $\rightarrow$ Monotone bounded convergence $\rightarrow$ Bolzano-Weierstrass $\rightarrow$ EVT $\rightarrow$ IVT $\rightarrow$ Rolle's Theorem $\rightarrow$ MVT, Cauchy's MVT $\rightarrow$ Integral MVT
Each theorem in the chain is useful by itself, but the main use of EVT is to act as a stepping stone in this chain.