 Solution

A real function on a metric space is uniformly continuous if for every $\epsilon >0$ there is a $\delta >0$ such that, simultaneously for all $x,y$, $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$. For a continuous function on a finite closed interval $[a,b]$, suppose this fails. Then there are ${\epsilon }_0>0$ and pairs $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. By compactness, some subsequence $x_{n_j}$ converges to $x\in [a,b]$; then $y_{n_j}\to x$ too. Continuity contradicts the image separation. This proves the Heine-Cantor theorem in this setting. Boundedness of the interval matters: $x^2$ on $[0,\infty )$ is a counterexample if “closed interval” includes unbounded intervals.

A function is Lipschitz continuous if some finite $K\ge 0$ satisfies $|f(x)-f(y)|\le Kd(x,y)$ for every pair. Choosing $\delta =\epsilon /(K+1)$ proves uniform continuity.

The first assertion is false. Take $f(x)=\sin (x^2)$, which is bounded and continuous, andThen $f(x_n)=1$, $f(y_n)=-1$, while $y_n-x_n=\pi /(y_n+x_n)\to 0$. This violates uniform continuity.