Let $\lambda$ be any real number and $f:[0,1]\times[0,1]\rightarrow\mathbb{R}$ given by $$ f(x,y) = \begin{cases} \lambda &\mbox{if } \quad0<x<1, y=1, \\ 1+\lambda y & \mbox{if } \quad\text{otherwise}. \end{cases} $$ Proving that $f$ is quasiconvex, i.e. $$ f[\alpha(x_1, x_2)+(1-\alpha)(y_1, y_2)]\leq \max\{f(x_1, x_2), f(y_1, y_2)\} $$ for all $(x_1, x_2), (y_1, y_2)\in[0,1]\times[0,1]$ and $\alpha\in (0,1)$.
My attemption. I think that we have to use the convexity of the lower level set of $f$ the deduce the quasiconvexity of $f$. Indeed, for $\lambda=0$, we have $$ f(x,y) = \begin{cases} 0 &\mbox{if } \quad 0<x<1, y=1, \\ 1 & \mbox{if } \quad\text{otherwise}. \end{cases} $$ Let $\alpha$ be a real number. The lower level set of $f$ is given $$ L(\alpha; f):=\{(x,y)\in[0,1]\times[0,1]: f(x, y)\leq \alpha\}= \begin{cases} [0,1]\times[0,1] &\mbox{if } \quad \alpha\geq 1, \\ (0,1)\times\{1\} & \mbox{if } \quad 0\leq\alpha<1,\\ \emptyset & \mbox{if } \quad \alpha<0.\end{cases} $$ Hence, $L(\alpha; f)$ is convex for all $\alpha\in\mathbb{R}$. It follows that $f$ is quasiconvex for $\lambda=0$.
I cannot check the quasiconvexity of $f$ for the case $\lambda\ne 0$.
For $\lambda=1$ we have $$ f(x,y) = \begin{cases} 1 &\mbox{if } \quad0<x<1, y=1, \\ 1+y & \mbox{if } \quad\text{otherwise}\end{cases} $$ and $$ L(3/2;f)=(0,1)\times\{1\}\bigcup [0,1]\times[0,1/2]. $$ Since $L(3/2;f)$ is not convex, $f$ is not quasiconvex on $C$ for $\lambda=1$.