Can Non-Compact Closed Subsets Be Disjoint in a Metric Space?

[Kreizhn]

Homework Statement

Give an example of a metric space $(X,d)$, and nonempty subsets A,B of X such that both A,B are closed, non-compact, disjoint ($A \cap B = \emptyset$), and $\forall k>0, \; \exists a \in A, b \in B$ such that d(a,b)<k

The Attempt at a Solution

I've been trying to consider the set of all infinite binary sequences
$$X = \left\{ (x^{(1)}, x^{(2)}, \ldots, x^{(n)}, \ldots ) | x^{(i)} \in \{0,1\} \forall i \geq 1 \right\}$$

but I ended up showing that this is a compact metric space and as such all closed subsets are necessarily compact.

So I'm not terribly sure about any other examples that might work...

Edit: X is a compact metric space under

$$d(x,y) = \displaystyle \sum_{k=1}^\infty \frac{1}{2^k} | x^{(k)} - y^{(k)} |$$

 

[Dick]

R^2 is not compact. And graphs of continuous functions R->R are closed. Does that suggest anything?