Let $g_n:=\sqrt{f_n}$ and $g:=\sqrt{f}$. Let 
$$v:=\|g\|,
$$
where $\|h\|:=\|\,h|_J\,\|_{L^2(J)}$ for $h\in L^2(\mathbb R^2)$, $J:=I^2$, $I:=[-u,u]$, and $u\in(0,1/20)$ is small enough so that 
$$v>u/10;
$$
such a number $u$ exists, because $g(0,0)^2=1/(2\pi)>1/400$ and $g$ is continuous. 

One of the following two cases must occur. 

*Case 1: $\|g_n\|\le\|g\|/2$*. Then 
$$\|\sqrt{f_n}-\sqrt f\,\|_{H^1}=\|g_n-g\|_{H^1}\ge\|g_n-g\|\ge\|g\|/2=v/2. 
$$
So, Case 1 is good, because $v$ is a positive universal constant.

*Case 2: $\|g_n\|>v/2$*. In this case, use the condition $g_n(x,x)\equiv0$ to note that for all $x$ and $y$ in $I$ 
$$g_n(x,y)^2\le\Big(\int_I|(D_2g_n)(x,z)|\,dz\Big)^2
\le\int_I(D_2g_n)(x,z)^2\,dz,
$$
where $D_2$ is the partial derivative wrt the second argument. So, 
$$\frac{v^2}4<\|g_n\|^2=\int_{I^2}g_n^2
\le\int_{I^3}dx\,dy\,dz\,(D_2g_n)(x,z)^2\le\|D_2g_n\|^2, 
$$
so that 
$$\|D_2g_n\|>v/2.
$$
On the other hand,
$$\|D_2g\|^2\le\frac1{2\pi}\,\int_{I^2}y^2\,dx\,dy<u^4/4<(u/40)^2<(v/4)^2,
$$
whence $\|D_2g\|<v/4$. So,
$$\|\sqrt{f_n}-\sqrt f\,\|_{H^1}=\|g_n-g\|_{H^1}\ge\|D_2g_n-D_2g\|\ge v/2-v/4=v/4. 
$$
So, Case 2 is good as well. $\Box$