Statement: Let $\mathbb{R}^{\omega}$ be the countably infinite product of $\mathbb{R}$ with itself, and let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all sequences that have only finitely many nonzero elements. Then, in the box topology, $\mathbb{R}^{\infty}$ is closed.
I have found a proof of such a fact, but I am having trouble understanding how we can draw the conclusion in the last step.
Proof: Let $x \in \mathbb{R}^{\omega} - \mathbb{R}^{\infty}$. Then, $x$ has only finitely many elements equal to zero. A neighbourhood of this sequence $x$ is given by $U = U_1 \times U_2 \times \ldots$, where $U_k = \mathbb{R}$ if $x_k = 0$ and $U_k = (0,2x_k)$ is $x_k > 0$ or $U_k = (2x_k,0)$ if $x_k < 0$. Then, $U$ contains no elements of $\mathbb{R}^{\infty}$, and $\mathbb{R}^{\infty}$ is closed.
I have bolded the part which I do not understand why it is true. If anyone can explain it to me, that would be a great help.
To show $\mathbb{R}^\infty$ is closed, it suffices to show the complement is open. They do so by showing that for any point in the complement, there is an open neighborhood of the point contained entirely in the complement.