If you know about the complex interpolation, the following proof will work. In fact, this proof is a standard way to prove Clarkson's inequality, which is an $L^p$-generalization of this inequality.
The inequality is equivalent to
$$\left[\left|\frac{x+y}{2}\right|^q+\left|\frac{x-y}{2}\right|^q \right]^{1/q}\leq\left[\frac{|x|^p+|y|^p}{2} \right]^{1/p} \tag{1}$$
for $1<p<2$. So let us begin by first proving this inequality for the extremal cases $(p,q)=(1,\infty)$ and $(2,2)$. The first case follows from the triangle inequality:
$$\left|\frac{x\pm y}{2}\right|\leq \frac{|x|+|y|}{2}, $$
and the second case is just the parallelogram law. Now the general case will follow by interpolating these two results. Define
$$ T(x,y)=\left(\frac{x+y}{2},\frac{x-y}{2} \right)\qquad (x,y\in\mathbb{C}). $$
Then the above estimates say that $T: \ell^q(\{1,2\})\to\ell^p_\nu\{(1,2)\}$ is bounded for $p=1$ and $p=2$ with operator norm $1$; here, $\ell^p_\nu$ denotes the $\ell^p$ space equipped with the measure $\nu(\{1\})=\nu(\{2\})=1/2$. It follows from the complex interpolation that $T:\ell^q(\{1,2\})\to\ell^p_\nu\{(1,2)\}$ is bounded and has norm $1$ for every $1\leq p\leq 2$, which proves the assertion.
By the way, since we are dealing with finite sequence spaces $\ell^p(\{1,2\})$, I think that there might be an elementary interpolation argument which avoids using the Riesz-Thorin theorem, but I'm not sure. (A direct application of M. Riesz's convexity theorem would work, of course, but it seems no big different to appealing to Riesz-Thorin.)
