Shortest distance between curves is always along common normal.

Let $P\left(x_0, y_0\right)$ be any point on the parabola $x^2=2y$.

Clearly, slope of line $x-y=1$ is $1$. 

Hence, slope of perpendicular line is $-1$.

${\frac{dy}{dx}}_P=$Slope of tangent at $P$ 

Therefore, $\frac{-1}{{\frac{dy}{dx}}_P}=$Slope of normal$=-1$

$\Rightarrow {\frac{dy}{dx}}_P=1$

$\Rightarrow x_0=1 \therefore y_0=\frac{1}{2}$

$\Rightarrow P\left(1,\frac{1}{2}\right)$

$\therefore$ Shortest distance $=\left|\frac{1-\frac{1}{2}-1}{\sqrt{1^2+1^2}}\right|=\frac{1}{2\sqrt{2}}$

option $\left(3\right)$