= Solution
A function $f:E\to\mathbb R$ has <uniform continuity> if for every $\varepsilon>0$ there is a $\delta>0$ such that, for all $x,y\in E$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The same $\delta$ must work everywhere in $E$.
\b[The product need not be uniformly continuous.] Both $f(x)=x$ and $g(x)=x$ are <Lipschitz continuous>, hence uniformly continuous on $\mathbb R$, but their product is $x^2$. For $x_n=n$ and $y_n=n+1/n$, the input separation tends to zero while
$$
|y_n^2-x_n^2|=2+\frac1{n^2}\ge2.
$$
This contradicts <uniform continuity>. The issue is captured by <products of unbounded uniformly continuous functions>; bounded factors would give a uniform product estimate.
\b[A uniformly continuous function on $(0,1)$ is bounded.] Take $\delta$ for output tolerance one and cover $(0,1)$ by finitely many intervals of radius less than $\delta$, centred at points $c_1,\ldots,c_m$ in $(0,1)$. Every $x$ is within $\delta$ of some $c_j$, so $|f(x)|\le1+\max_j|f(c_j)|$. This proves the general principle that a <uniformly continuous function on a totally bounded set is bounded>.
\b[The reciprocal-cosine function is not uniformly continuous.] Take
$$
x_n=\frac1{2\pi n},\qquad y_n=\frac1{(2n+1)\pi}.
$$
Both sequences lie in $(0,1)$, $|x_n-y_n|\to0$, but the two function values are $1$ and $-1$. Their difference is always two, violating the <sequential criterion for uniform continuity>.
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