= Solution
We prove Booleanity directly from the dense-cover description. Let $B$ be a subsheaf of a sheaf $F$. Its pseudocomplement $N$ consists of sections $x\in F(c)$ such that no restriction along any arrow into $c$ belongs to $B$. This is a subpresheaf. It is also a sheaf: if a section of $F$ restricts into $N$ on a dense covering <sieve on a category>, any hypothetical restriction into $B$ has a further restriction in that cover; it would then belong to both $B$ and $N$, impossible. Thus the unique amalgamation in $F$ of an $N$-matching family lies in $N$.
For any $x\in F(c)$, the <sieve on a category> of arrows on which its restriction lies in $B$ or $N$ is dense. Given an arrow $u:d\to c$, either some further restriction belongs to $B$, supplying the required member of that <sieve on a category>, or none does, in which case the restriction along $u$ itself belongs to $N$. Therefore $B\cup N$ is locally all of $F$, so their join in the <sheaf topos> is $F$. Their intersection is empty. Every subsheaf consequently has a complement, proving \b[the topos is Boolean].
For two-valuedness, let $U\hookrightarrow1$ be a nonempty subterminal sheaf. At some stage $m$, its value is the singleton. For every $k\ge m$ there is an arrow $k\to m$, so restriction makes $U(k)$ the singleton as well. At any stage $n$, the <sieve on a category> of arrows into $n$ whose domains are at least $m$ is dense: after an arrow with domain $l$, choose a further domain $k\ge\max(l,m)$ and use the existence of arrows downwards. On this covering <sieve on a category> the unique terminal section has a matching family in $U$, so the sheaf condition gives $U(n)=1$. Hence $U=1$. Since the empty <sieve on a category> never covers, the empty presheaf is an initial sheaf distinct from $1$. Thus
$$
\boxed{\operatorname{Sub}(1)=\{0,1\},}
$$
and \b[the topos is two-valued], in the distinct global sense of a <two-valued topos>.
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