= Solution
A <function> $f:I\to\mathbb R$ is <uniformly continuous> if
$$
\forall\varepsilon>0\ \exists\delta>0\ \forall x,y\in I:
\quad |x-y|<\delta\Longrightarrow |f(x)-f(y)|<\varepsilon.
$$
The same $\delta$ must work at every point of $I$.
\b[Every <uniformly continuous> <function> on $(0,1)$ is bounded.] Choose the tolerance $\delta$ for $\varepsilon=1$ and an integer $N$ with $1/N<\delta$. The finitely many centers $c_j=(j+1/2)/N$, $0\leq j<N$, cover the interval by cells of radius $1/(2N)$. For each $x$, one center satisfies $|f(x)-f(c_j)|<1$. Thus $|f(x)|\leq1+\max_j|f(c_j)|$. This is the <uniformly continuous function on a totally bounded set is bounded> argument; closedness of the interval is unnecessary.
\b[The reciprocal <function> is not <uniformly continuous> on $(0,1)$.] It is unbounded, already contradicting the preceding result. Directly, $x_n=1/(n+1)$ and $y_n=1/(n+2)$ have distance tending to zero while $|1/x_n-1/y_n|=1$.
\b[The oscillating <function> $\sin(1/x)$ is not <uniformly continuous> either], despite being bounded. Take
$$
x_n=\frac1{2\pi n+\pi/2},\qquad y_n=\frac1{2\pi n+3\pi/2}.
$$
Their distance tends to zero, but their <function> values are $1$ and $-1$. The fixed output gap of two contradicts the uniform-continuity condition, for example with $\varepsilon=1$.
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