= Solution
\b[No: there need not be even one orbit tending to zero.] Partition the domain into the disjoint intervals
$$
I_j=\left(\frac1{j+1},\frac1j\right]\cap(0,1),\qquad j=1,2,\ldots,
$$
and, for the unique $j$ with $x\in I_j$, define
$$
\boxed{f(x)=\frac12\left(x+\frac1{j+1}\right).}
$$
The left endpoint is strictly below $x$, so $0<f(x)<x$. Moreover $f(x)>1/(j+1)$ and $f(x)\leq x\leq1/j$, so $f(x)$ remains in the same interval. For every iterate,
$$
\boxed{f^n(x)=\frac1{j+1}+2^{-n}\left(x-\frac1{j+1}\right)\longrightarrow\frac1{j+1}>0.}
$$
Every starting point belongs to some finite-index interval, so this proves the claim for all $x\in(0,1)$. The open left endpoints matter: an orbit approaches a boundary but never crosses it. This is <discontinuous trapping of decreasing iterates>, not a violation of the <bounded monotone sequence theorem>; each orbit does converge, just to a positive value where continuity fails.
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