Mapping whose Graph is Closed in Chebyshev Product is not necessarily Continuous

Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $\AA = A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.

Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
 * $\ds \map {d_\infty} {x, y} = \max \set {\map {d_1} {x_1, y_1}, \map {d_2} {x_2, y_2} }$

where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.

Let $\Gamma_f$ be the graph of $f$.

Let $f: M_1 \to M_2$ be a mapping such that the $\Gamma_f$ is a closed set of $\struct {A_1 \times A_2, d}$.

Then it is not necessarily the case that $f$ is a continuous mapping.