Is $[1,2]\times \Bbb R$ closed? Is $[1,2]\times \Bbb R$ closed in $\Bbb R^2$? The answer is no. But I suggest it is yes, since it contains all its boundary points right? Am I right or wrong?

Take a sequence $(x_n)$ of points in $[1,2]\times \mathbb{R}$. If we write $x_n=(y_n,z_n)$ in the coordinates on the product space then we  see that if the sequence $x_n$  converges, then so do the  sequences $(y_n)$ and $(z_n)$ in $[1,2]$ and $\mathbb{R}$, respectively. Now, $[1,2]$ and $\mathbb{R}$ are closed so that $y_n\to y\in [1,2]$ and $z_n\to z\in \mathbb{R}$. So, $x_n\to x=(y,z)\in [1,2]\times \mathbb{R}$. Indeed, $[1,2]\times \mathbb{R}$  is closed. 
Remark: As mentioned in the comments: $[1,2]\times \mathbb{R}$ is not compact because it is not bounded. The Heine-Borel Theorem says that in the topology on $\mathbb{R}^2$, $\Omega\subseteq \mathbb{R}^2$ is compact if and only if it is closed and bounded.
