Solution:
First, given $a$, let $C_{a}$ be the set of all $b \in A$ for which there exists a sequence $a=c_{0}, c_{1}, \ldots, c_{n}=b$ as in the definition of a cell. Certainly $a \in C_{a}$ (take $n=0$); we claim that $C_{a}$ is a cell. If $b, b^{\prime} \in C_{a}$, then there exist sequences $a=c_{0}, c_{1}, \ldots, c_{n}=b$ and $a=c_{0}^{\prime}, c_{1}^{\prime}, \ldots, c_{m}^{\prime}=b^{\prime}$, so the sequence $b=c_{n}, c_{n-1}, \ldots, c_{1}, c_{0}, c_{1}^{\prime}, \ldots, c_{m}^{\prime}=b^{\prime}$ shows that the first condition is met.

For the second, suppose that there does exist a sequence $b=c_{0}, c_{1}, \ldots, c_{n}=b^{\prime}$ with $b \in C_{a}, b^{\prime} \notin C_{a}$. Then, concatenating with our sequence from $a$ to $b$, we get a sequence from $a$ to $b^{\prime}$, contradicting the assumption $b^{\prime} \notin C_{a}$. Thus, the second condition holds, and $C_{a}$ is a cell. So $a$ lies in at least one cell.

But now, note that if $C$ is a cell containing $a$, then all $b$ for which such a sequence from $a$ to $b$ exists must lie in $C$ (or the second condition is violated), and if no such sequence exists, then $b$ cannot lie in $C$ (or the first condition is violated). Thus, the elements of $C$ are uniquely determined, so there is exactly one cell containing $a$, and the proof is complete.