[proofplan] We partition Hechler forcing according to the finite stem of a condition. There are only countably many finite stems, so this gives a countable cover once each piece is shown to be centered. Conditions with the same stem have a common extension obtained by taking the pointwise maximum of their bounding functions. Finally, every antichain meets each centered piece in at most one condition, so every antichain is countable. [/proofplan] [step:Cover $\mathbb{D}$ by the conditions with a fixed finite stem] For each finite sequence $s \in \omega^{<\omega}$, define $C_s := \{(s,f) \in \mathbb{D} : f \in \omega^\omega\}$. Every condition $(t,h) \in \mathbb{D}$ belongs to $C_t$, so $\mathbb{D} = \bigcup_{s \in \omega^{<\omega}} C_s$. The set $\omega^{<\omega}$ is countable because $\omega^{<\omega} = \bigcup_{m \in \omega} \omega^m$, and each finite Cartesian power $\omega^m$ is countable. Hence the cover above is a countable cover of $\mathbb{D}$. [/step] [step:Show that each fixed-stem piece is centered] Fix $s \in \omega^{<\omega}$. We prove that $C_s$ is centered. The empty finite subset of $C_s$ has a common extension by the usual convention for centered families, so it remains to consider nonempty finite subsets. Let $k \in \omega$ with $k \geq 1$, and let $(s,f_0), \dots, (s,f_{k-1}) \in C_s$ be finitely many conditions. Define the function $g: \omega \to \omega$ by $g(n) := \max\{f_i(n) : i < k\}$ for each $n \in \omega$. Then $g \in \omega^\omega$, so $(s,g) \in \mathbb{D}$. For each $i < k$, we have $s \subseteq s$, $g(n) \geq f_i(n)$ for every $n \in \omega$, and $\operatorname{dom}(s) \setminus \operatorname{dom}(s) = \varnothing$. Thus the third clause in the order definition is vacuous, and therefore $(s,g) \leq (s,f_i)$ for every $i < k$. Hence every finite subset of $C_s$ has a common extension, so $C_s$ is centered. [guided] Fix a finite sequence $s \in \omega^{<\omega}$. To show that $C_s$ is centered, we must prove that every finite family of conditions in $C_s$ has a common extension in the forcing order. The empty finite subset has a common extension by the usual convention in the definition of centeredness, so we consider a nonempty finite family. Let $k \in \omega$ with $k \geq 1$, and choose finitely many conditions $(s,f_0), \dots, (s,f_{k-1}) \in C_s$. All these conditions have the same stem $s$; only their second coordinates may differ. The natural way to strengthen all of them at once is to choose a single bounding function that dominates every $f_i$ pointwise. Define the function $g: \omega \to \omega$ by $g(n) := \max\{f_i(n) : i < k\}$ for each $n \in \omega$. Since this maximum is taken over a nonempty finite subset of $\omega$, it exists and lies in $\omega$. Therefore $g \in \omega^\omega$, and hence $(s,g)$ is a condition in $\mathbb{D}$. We now verify the forcing order. Fix $i < k$. First, the stem of $(s,g)$ extends the stem of $(s,f_i)$ because $s \subseteq s$. Second, by the definition of $g$, $g(n) \geq f_i(n)$ for every $n \in \omega$. Third, there are no new stem coordinates to check, since $\operatorname{dom}(s) \setminus \operatorname{dom}(s) = \varnothing$. Thus every clause in the definition of the Hechler order is satisfied, so $(s,g) \leq (s,f_i)$. Because this holds for each $i < k$, the condition $(s,g)$ is a common extension of the chosen finite family. Therefore $C_s$ is centered. [/guided] [/step] [step:Conclude that $\mathbb{D}$ is $\sigma$-centered] The family $\{C_s : s \in \omega^{<\omega}\}$ is countable, covers $\mathbb{D}$, and each $C_s$ is centered. Therefore $\mathbb{D}$ is $\sigma$-centered. [/step] [step:Deduce the countable chain condition] Let $A \subseteq \mathbb{D}$ be an antichain. If $p,q \in A \cap C_s$ are distinct, then $p$ and $q$ have a common extension because $C_s$ is centered, contradicting that $A$ is an antichain. Hence $A \cap C_s$ has at most one element for each $s \in \omega^{<\omega}$. Since $A \subseteq \bigcup_{s \in \omega^{<\omega}} (A \cap C_s)$ and $\omega^{<\omega}$ is countable, the antichain $A$ is countable. Thus $\mathbb{D}$ satisfies the countable chain condition. [/step]