Free band on $n$ generators is a semilattice of rectangular bands. Exercise (part of it) comes from Clifford, Preston, Algebraic theory of Semigroups.
By $\mathbb{FB}_n$ we denote the free band on $n$ generators, that is, a free semigroup generated by a set $X=\{x_1,...,x_n\}$ subject to relations $w^2 = w$ where $w$ is an arbitrary word in the free semigroup on $X$. 
Denote by $Y$ the semilattice of subsets of $X$ under union of sets (which is isomorphic to the free commutative band on $X$).
Part b) of the exercise tells me that $\mathbb{FB}_n$ is a semilattice $Y$ of rectangular bands $S_\alpha$, $\alpha\in Y$, where $S_\alpha$ consists of all words $w$ such that the set of generators appearing in $w$ is $\alpha$. 
From my understanding, for example $\alpha = \{x_1, x_2\}$, then $S_\alpha$ is the free band generated by $\alpha$. But this is not a rectangular band, since $x_1x_2x_1 \neq x_1$. 
My question is, what exactly do they mean by $S_\alpha$?
 A: Let me first recall a few definitions. A band is an idempotent semigroup
and a semilattice is an idempotent and commutative semigroup. A semilattice congruence on a semigroup $S$ is a congruence $\equiv$ on $S$ such that the quotient semigroup $S/{\equiv}$ is a semilattice. The minimum semilattice congruence on $S$ is the intersection of all semilattice congruences on $S$. Alternatively, it is the congruence generated by the relations $s \sim s^2$ and $st \sim ts$ for all $s, t \in S$. Finally,
two elements of a semigroup are
$\mathcal{J}$-related if they generate the same ideal.
The following results hold:

Proposition 1. Let $S$ be a band, let $T$ be a semilattice and let $h:S \to T$ be a semigroup morphism. If $a \mathrel{\mathcal{J}} b$, then $h(a) = h(b)$.

Proof. Since a semigroup morphism preserves $\mathcal{J}$-classes, $h(a) \mathrel{\mathcal{J}} h(b)$. But in a semilattice, the $\mathcal{J}$-relation is the equality. Thus $h(a) = h(b)$.

Proposition 2. Let $S$ be a band. Then the $\mathcal{J}$-relation
  is a congruence on $S$ and the quotient $S/{\mathcal{J}}$ is a semilattice.

Proof. Let $a, b \in S$. Then $ab = (ab)^2 = a(ba)b$ and $ba = (ba)^2 = b(ab)b$. Thus $ab \mathrel{\mathcal{J}} ba$. Suppose now that $a \mathrel{\mathcal{J}} b$ and let $c \in S$. Then $b = xay$ and $a = ubv$ for some $x,y,u,v \in S^1$. Therefore
$$
ca = c(ubv) = (cu)bv \leqslant_\mathcal{J} (cu)b \mathrel{\mathcal{J}} b(cu) \leqslant_\mathcal{J} bc \mathrel{\mathcal{J}} cb
$$
and similarly, $cb \leqslant_\mathcal{J} ca$, whence $cb \mathrel{\mathcal{J}} ca$. It follows that $\mathcal{J}$ is a left congruence and a dual argument would show that it is a right congruence. Finally, $S/{\mathcal{J}}$ is idempotent and since $ab \mathrel{\mathcal{J}} ba$, it is also commutative. 

Corollary 1. In a band, the minimum semilattice congruence is equal to  $\mathcal{J}$.

Proof. Let $S$ be a band, let $\sim$ be its minimum semilattice congruence and let $p:S \to S/{\sim}$ be the quotient morphism. Proposition 1 shows that if $a \mathrel{\mathcal{J}} b$, then $p(a) = p(b)$, whence $a \sim b$. Moreover, Proposition 2 shows that $\mathcal{J}$ is a semilattice congruence and hence contains $\sim$. Thus $a \sim b$ implies $a \mathrel{\mathcal{J}} b$, which concludes the proof.

Corollary 2. In $\mathbb{FB}_n$, the $\mathcal{J}$-relation is the minimum semilattice congruence and the quotient $\mathbb{FB}_n/{\mathcal{J}}$ is the free semilattice $\mathbb{FSl}_n$. 

Back to your question. The free semilattice $\mathbb{FSl}_n$ on $X$ can be identified to the semigroup $(\mathcal{P}(X), \cup)$, where $\mathcal{P}(X)$ is the set of nonempty subsets of $X$. Consider the following quotient morphisms
$$
  X^+ \xrightarrow{q} \mathbb{FB}_n \xrightarrow{p} \mathbb{FSl}_n = (\mathcal{P}(X), \cup) \quad \text{and let} \quad c = p \circ q: X^+ \to (\mathcal{P}(X), \cup)
$$
Then for each word $w \in X^+$, $c(w)$ is the content of $w$, that is, the set of letters occurring in $w$. Since $c(uv) = c(u) \cup c(v)$, the content map is indeed a semigroup morphism from $X^+$ to $(\mathcal{P}(X), \cup)$. 
Furthermore, $q(u) \mathrel{\mathcal{J}} q(v)$ if and only if $c(u) = c(v)$.
Finally, for each subset $Y$ of $X$, $p^{-1}(Y)$ is the $\mathcal{J}$-class
consisting of the elements of the form $q(u)$, where $c(u) = Y$.
It is a rectangular band, and one of the $S_\alpha$ of your question. 
Example. If $X= \{a,b\}$, there are three $\mathcal{J}$-classes in the free band on $X$: $\{a\}$, $\{b\}$ and $\{ab, aba, bab, ba\}$ (the elements of content $\{a,b\}$. You can verify by hand that each of them is a rectangular band.
