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For all $x$ and $y$ in $[0,1]^2$ $$f(x,y)= \left\{ \begin{aligned} \int_x^y f_y(x,z)\,dz&\text{ if }x\le y, \\ -\int_y^x f_y(x,z)\,dz&\text{ if }x\ge y, \end{aligned} \right. $$ where $f_y(x,z):=\frac{\partial f(x,y)}{\partial y}|_{y=z}$, so that
$$|f(x,y)|\le\int_0^1|f_y(x,z)|\,dz\le\sqrt{\int_0^1|f_y(x,z)|^2\,dz}. $$ Hence, $$ \begin{split} 1 & = \iint_{[0,1]^2}dx\,dy\,|f(x,y)|^2 \\ & \le \iiint_{[0,1]^3}dx\,dy\,dz\,|f_y(x,z)|^2\\ & = \iint_{[0,1]^2}dx\,dz\,|f_y(x,z)|^2 \le\|f-1\|_{H^1}^2. \end{split} $$ So, any $c\in(0,1)$ will do.