Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 2 / d / ii / Solution

If $f$ is continuous at $a$, then for every $\epsilon >0$ there is $r_0>0$ such that $d(x,a)<r_0$ implies $|f(x)-f(a)|<\epsilon$. In particular, for $0<r\le r_0$,The squeeze theorem giveswhich implies condition (1). Thus continuity implies upper semicontinuity.