Sorry for the silly question. I just wanted to confirm whether the
Lipschitz constant of gradient of $f(z):= \frac{1}{2}\|y - z\|_2^2$ over $\mathbb{R}^n$ is 1, isn't it?
The definition of Lipschitz continuous gradient is \begin{align} \| \nabla f(z_1) - \nabla f(z_2)\|_2 \leq L \|z_1 - z_2\|_2 \quad \forall z_1, z_2 \in \mathbb{R}^n. \end{align}
The gradient of $f(z)$ is $\nabla f(z) = -(y-z)$. So, plugging this in the above definition, \begin{align} & \| \underbrace{\nabla f(z_1)}_{-y+z_1} - \underbrace{\nabla f(z_2)}_{-y+z_2}\|_2 \leq L \|z_1 - z_2\|_2 \\ \Longleftrightarrow & \| z_1 - z_2 \| \leq L \|z_1 - z_2 \| \quad \Rightarrow \quad L \geq 1. \end{align}
ADD: Of course $L$ may depend on the considered norm. I have assumed L2 norm though.
Certainly $\nabla f$ is non-expansive, by the argument provided. The argument doesn't quite rule out the possibility of some $L < 1$ also satisfying the condition (though it comes very close to concluding this), so the Lipschitz constant can only be concluded to be less than or equal to $1$.
To show the Lipschitz constant is indeed $1$, consider $z_1 = (1, 0) + y$ and $z_2 = y$. Then, for any eligible $L$, $$\|\nabla f(z_1) - \nabla(z_2)\| \le L\|z_1 - z_2\| \implies \|(1, 0) \| \le L\|(1, 0)\| \implies L \ge 1.$$ Thus the Lipschitz constant is $1$.