Edit.

Let us first prove the inequality:
$$\sup_{z,w}|f(z)-f(w)|\geq\sup_z|f(z)-f(-z)|=\sup|g(z)|\geq|g'(0)|=2|f'(0)|,$$
where the Schwarz Lemma was applied to $g(z)=f(z)-f(-z)$. Equality in Schwarz lemma can
happen only if $g(z)=kz$, thus $f(z)=kz+\phi(z)$, where $\phi$ is even.

Now let us see when equality is possible in the first inequality. We must have
$$\sup_{|z|<1,|w|<1}|k(z-w)+\phi(z)-\phi(w)|=2|k|,$$
which an even function $\phi$ analytic in the unit disc. We have to derive from here that
$\phi$ is constant.

In the reference
http://www.math.wustl.edu/~geknese/schwarzpoly.pdf,
where extremal functions for the Schwarz lemma in the polydisc are described.
But those functions are extremal at every point.
And our function is extremal at only one point, the origin.

I can prove that $\phi$ is constant only under the additonal restriction that
$\phi$ is differentiable in the closed disc. WLOG $k=1$.
Let $|z|=1$ and put $w=-ze^{it}$, where $t$ is small. Then, neglecting the high powers
of $t$, we must have
$$|2+it-\phi'(z)it|\leq 2$$
This implies that $\phi'(z)$ must be real for all $z$ on the unit circle. But such function must be constant, and as $\phi$ is even, we conclude that $\phi$ is constant.

To get rid of the additional assumption of differentiability, one may combine
this and
and this.