how to prove that sigma-compact space is D-space

In the paper "A survey of D-spaces" by Gary Gruenhage it is written that it is easily seen that $\sigma$-compact spaces are D-spaces.

Unfortunately, I don't know how to show it. The considered spaces are regular and $T_1$. Thank you for your help! Karel Pastor

(Added by Asaf) The definition of a D-space, taken from the abstract of the aforementioned paper:

A space $X$ is a D-space if whenever one is given a neighborhood $N(x)$ of $x$ for each $x \in X$, then there is a closed discrete subset $D$ of $X$ such that $\{N(x) \colon x \in D\}$ covers $X$.

-

I think this works: pick an increasing family of compact sets $K_n$, which covers $X$. Because of compactness, a finite subfamily of $\{N(x);x\in K_0\}$ covers $K_0$; denote their "base points" by $x^0_0,\ldots,x^0_{n_0}$. Now, by compactness again, a finite subfamily of $\{N(x)\cap K_0^c;x\in K_1\setminus\bigcup_{i=0}^{n_0}N(x^0_i)\}$ covers $K_1\setminus\bigcup_{i=0}^{n_0}N(x^0_i)$; denote their base points by $x^1_0,\ldots,x^1_{n_1}$. Continuing in this way, we get a set $D=\bigcup_{i=0}^\infty\{x^i_{0},\ldots,x^i_{n_i}\}$, the neighborhoods of whose points cover $X$.

Now pick a point $x^i_j\in D$. Using Hausdorffness, we can separate $x^i_j$ from all of the points in $D\cap K_i$. Since we picked the points $x^k_l, k>i,$ away from $N(x^i_j)$, we can separate $x^i_j$ from the points $(D\cap K_i)^c$. Therefore $x^i_j$ is an isolated point in $D$. This proves the discreteness of $D$. Almost the same argument shows that $D$ is closed in $X$.

-
Thank you very much! Karel Pastor –  Karel Pastor Aug 15 '11 at 19:14