inequality involving complex exponential

One way is to use $$ |e^{ix} - e^{iy}| = \left|\int_x^ye^{it}\,dt\right|\leq \int_x^y\,dt = y-x, $$ assuming $y > x$.

A graphical hint (apparently I need 30 characters... doesn't stackexchange know that a picture is worth a thousand words?).

The function $u$ defined on $[0,1]$ by $u(t)=\exp(\mathrm i x+\mathrm it(y-x))$ is such that $u'(t)=\mathrm i(y-x)u(t)$ and $|u(t)|=1$ hence $|u'(t)|=|y-x|$ and $|u(1)-u(0)|\leqslant\sup\{|u'(t)|;t\in[0,1]\}=|y-x|$. Note finally that $u(0)=\exp(\mathrm i x)$ and $u(1)=\exp(\mathrm i y)$.

Hints: $e^{ix} -e^{iy}= e^{ix}(1 -e^{i(y-x)})$, and $|e^{ix}|=1$. Let $\theta=x-y$.

Hint: $$\begin{align}|e^{ix}-e^{iy}|^2 &= (\cos x-\cos y)^2 + (\sin x - \sin y)^2 \\ &= 2-2(\sin x \sin y +\cos x \cos y) = 2-2\cos (x-y)\end{align}$$

For the last step, see ProofWiki. Now use the series expansion of $\cos(x-y)$ and compare to $(x-y)^2$.