[proofplan] We prove the stronger assertion that every $\sigma$-linked forcing notion is ccc. Given a countable linked cover $\mathbb{P} = \bigcup_{n \in \omega} L_n$ and an antichain $A \subset \mathbb{P}$, assign each condition in $A$ to the first linked piece containing it. Two distinct members of $A$ cannot receive the same index, because that would make them compatible inside a linked set. Thus $A$ injects into $\omega$, so $A$ is countable. The $\sigma$-centered case follows because every centered subset is linked. [/proofplan] [step:Assign each antichain element to the first linked set containing it] Let $\omega$ denote the set of natural numbers, including $0$. Assume that $\mathbb{P}$ is $\sigma$-linked. Thus there is a sequence $(L_n)_{n \in \omega}$ of subsets of $\mathbb{P}$ such that each $L_n$ is linked and \begin{align*} \mathbb{P} = \bigcup_{n \in \omega} L_n. \end{align*} Let $A \subset \mathbb{P}$ be an antichain. Define a function \begin{align*} f: A \to \omega,\quad p \mapsto \min\{n \in \omega : p \in L_n\}. \end{align*} This is well-defined because every $p \in A$ belongs to $\mathbb{P}$, and the displayed union covers $\mathbb{P}$. [guided] Let $\omega$ denote the set of natural numbers, including $0$. We start from the definition of $\sigma$-linked. This means that $\mathbb{P}$ is covered by countably many linked pieces, so we choose a sequence $(L_n)_{n \in \omega}$ of subsets of $\mathbb{P}$ with each $L_n$ linked and \begin{align*} \mathbb{P} = \bigcup_{n \in \omega} L_n. \end{align*} Now let $A \subset \mathbb{P}$ be an antichain. To prove that $\mathbb{P}$ is ccc, it is enough to show that this arbitrary antichain $A$ is countable. The natural idea is to record where each condition of $A$ first appears in the countable cover. Define \begin{align*} f: A \to \omega,\quad p \mapsto \min\{n \in \omega : p \in L_n\}. \end{align*} For each $p \in A$, the set $\{n \in \omega : p \in L_n\}$ is nonempty because the $L_n$ cover all of $\mathbb{P}$ and $p \in \mathbb{P}$. Since every nonempty subset of $\omega$ has a least element, $f(p)$ is defined for every $p \in A$. [/guided] [/step] [step:Show that the first linked index map is injective] We claim that $f$ is injective. Let $p, q \in A$ and suppose $f(p) = f(q) = n$. Then $p \in L_n$ and $q \in L_n$. Since $L_n$ is linked, $p$ and $q$ are compatible. Since $A$ is an antichain, no two distinct elements of $A$ are compatible. Therefore $p = q$. Hence $f$ is injective. [/step] [step:Conclude that every antichain is countable] Since $f: A \to \omega$ is injective, the antichain $A$ is countable. Because $A \subset \mathbb{P}$ was arbitrary, every antichain in $\mathbb{P}$ is countable. Therefore $\mathbb{P}$ satisfies the countable chain condition. [/step] [step:Deduce the sigma-centered case from the sigma-linked case] Assume that $\mathbb{P}$ is $\sigma$-centered. Then there is a sequence $(C_n)_{n \in \omega}$ of subsets of $\mathbb{P}$ such that each $C_n$ is centered and \begin{align*} \mathbb{P} = \bigcup_{n \in \omega} C_n. \end{align*} Each centered set $C_n$ is linked: if $p, q \in C_n$, then the finite set $\{p, q\}$ has a common extension $r \in \mathbb{P}$ with $r \leq p$ and $r \leq q$, so $p$ and $q$ are compatible. Thus the same sequence $(C_n)_{n \in \omega}$ is a countable linked cover of $\mathbb{P}$, so $\mathbb{P}$ is $\sigma$-linked. By the result proved above, $\mathbb{P}$ is ccc. [/step]