\b[No.] Give an infinite set $X$ the <discrete metric>
$$
d(x,y)=\begin{cases}0,&x=y,\\1,&x\ne y.\end{cases}
$$
This space is bounded. Every Cauchy sequence is eventually constant, so it is complete. However, the open cover by the singleton balls $B_{1/2}(x)=\{x\}$ has no finite subcover. Hence $X$ is not <compact space>[compact].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 4 / 12E / a / ii / Solution

For a locally finite <discrete metric> space $(X,d)$ with basepoint $x_0$, the <growth function of a discrete metric space> is
$$
\beta_{X,x_0}(r)=|B(x_0,r)|
=\#\{x\in X:d(x,x_0)\leq r\}.
$$
In a <Cayley graph>, left multiplication by $gx_0^{-1}$ is a graph isometry carrying $x_0$ to any other vertex $g$. It therefore gives a bijection $B(x_0,r)\to B(g,r)$ for every $r$. Thus \b[the growth function of a fixed Cayley graph is exactly independent of the basepoint].

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 143 / 3 / a / Solution

{title2=$d(x,y)=\boldsymbol1_{\{x\ne y\}}$}
{wiki}

The discrete metric sets $d(x,x)=0$ and $d(x,y)=1$ for distinct points. It induces the <discrete topology>, and its Cauchy sequences are exactly the eventually constant sequences.

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