Projection on Real Euclidean Plane is not Closed Mapping

Theorem
Let $\left({\R^2, d}\right)$ be the real Euclidean plane.

Let $\rho: \R^2 \to \R$ be the first projection on $\R^2$ defined as:
 * $\forall \left({x, y}\right) \in \R^2: \rho \left({x, y}\right) = x$

Then $\rho$ is not a closed mapping.

The same applies with the second projection on $\R^2$.

Proof
Consider the set $A \subseteq R^2$ of all points defined as:
 * $A := \left\{{\left({x, y}\right) \in \R^2: x y \ge 1}\right\}$

By Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed:
 * $A$ is a closed set in $\left({\R^2, d}\right)$.

By inspection, it can be seen that the image of $A$ under $\rho$ is:
 * $\rho \left[{A}\right] = \left({\gets \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \to}\right)$

which by Union of Open Sets of Metric Space is Open is open.

Hence the result by definition of closed mapping.