On closedness: commodity space
A closed set does not need to be bounded. For instance, the set $[0,\infty)$ is closed but unbounded.

Formally, a set is closed if it contains all its limit points. You can easily verify that it is the case for your $C$. Take a sequence of elements $x^m=(x_1^m,\cdots,x_n^m) \in C$ that converges to a vector $x^{*}=(x_1^{*},\cdots,x_n^{*})$ under any appropriate topology. The $j$-th coordinate of $x^{*}$, $x_j^{*}$, is the limit of the sequence $x_j^{m}$. Since all $x_j^{m}$ are nonnegative real numbers, so is the case for the limit $x_j^{*}$. Since this is true for all $j$, $x^{*}$ also belongs to $C$.