Solution
= Solution
{c}
If $f$ is <continuous> at $a$, then for every $\varepsilon>0$ there is $r_0>0$ such that $d(x,a)<r_0$ implies $|f(x)-f(a)|<\varepsilon$. In particular, for $0<r\leq r_0$,
$$
f(a)\leq\sup_{x\in B(a,r)}f(x)\leq f(a)+\varepsilon.
$$
The <squeeze theorem> gives
$$
\boxed{\lim_{r\downarrow0}\sup_{x\in B(a,r)}f(x)=f(a),}
$$
which implies condition (1). Thus <continuity> implies <upper semicontinuity>.