= Solution
{c}
Suppose the <continuous function> $f$ did not have <uniform continuity> on $[a,b]$. Negating its definition gives $\varepsilon_0>0$ such that for every $n\geq1$ there are $x_n,y_n\in[a,b]$ with
$$
|x_n-y_n|<1/n,\qquad |f(x_n)-f(y_n)|\geq\varepsilon_0.
$$
The <Bolzano-Weierstrass theorem>, which states that every <bounded sequence> of <real numbers> has a <convergent subsequence>, gives $x_{n_k}\to p$. The inequalities $a\leq x_{n_k}\leq b$ imply $p\in[a,b]$. By the <triangle inequality>,
$$
|y_{n_k}-p|\leq|y_{n_k}-x_{n_k}|+|x_{n_k}-p|\longrightarrow0.
$$
Since <continuous functions preserve convergent sequences>, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its <absolute value> being at least $\varepsilon_0$. This proves the <Heine-Cantor theorem> in the required setting: \b[$f$ has <uniform continuity> on $[a,b]$.]
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