= Solution
The <Bonnet-Myers theorem> says that a <connected> <geodesically complete> $n$-dimensional <Riemannian manifold>, with $n\geq2$ and $\operatorname{Ric}\geq(n-1)\kappa g$ for a constant $\kappa>0$, has
$$
\boxed{\operatorname{diam}M\leq\frac{\pi}{\sqrt\kappa},\qquad M\text{ compact},\qquad |\pi_1(M)|<\infty}.
$$
The <dimension> and positive lower bound are essential: in <dimension> one the Ricci condition is vacuous, and a zero lower bound does not imply bounded diameter.
By Hopf-Rinow, any two distinct points have a unit-speed <minimizing geodesic> $\gamma:[0,L]\to M$. Choose parallel orthonormal fields $E_1,\ldots,E_{n-1}$ perpendicular to its tangent $T$. For the endpoint-vanishing fields $V_i(t)=\sin(\pi t/L)E_i(t)$, the <second variation of geodesic energy> gives nonnegative index forms
$$
I(V_i,V_i)=\int_0^L\bigl(|D_tV_i|^2-g(R(V_i,T)T,V_i)\bigr)\,dt\geq0.
$$
Summing and using the Ricci lower bound produces the <sine index-form bound for positive Ricci curvature>:
$$
0\leq\sum_iI(V_i,V_i)
\leq(n-1)\int_0^L\left[\frac{\pi^2}{L^2}\cos^2(\pi t/L)-\kappa\sin^2(\pi t/L)\right]dt
=\frac{n-1}{2}\left(\frac{\pi^2}{L}-\kappa L\right).
$$
If $L>\pi/\sqrt\kappa$, the last expression is negative, a contradiction. This proves the diameter bound. Hopf-Rinow makes closed bounded sets <compact>, so the entire manifold is <compact>.
Give the <universal cover> the pullback metric. <Local isometry> preserves its Ricci bound, and lifting complete base <geodesics> proves completeness of the cover. The same diameter and <compactness> argument applies there. A fiber of the covering is closed and discrete, hence finite in this <compact> cover; its cardinality is that of the <fundamental group>. This proves the final assertion.
For a counterexample that also breaks the diameter conclusion, use the <incomplete positively curved strip with infinite diameter>
$$
M=(-\pi/4,\pi/4)\times\mathbb R,\qquad g=du^2+\cos^2u\,dv^2.
$$
The map $\Phi(u,v)=(\cos u\cos v,\cos u\sin v,\sin u)$ is a <local isometry> to the round unit sphere: its coordinate <derivatives> are orthogonal, with squared lengths one and $\cos^2u$. Thus $K=1$ and $\operatorname{Ric}=g$, satisfying the required lower bound with $n=2$, $\kappa=1$.
The meridian $(u,v)=(t,0)$ is a unit-speed <geodesic> and reaches the excluded boundary at $t=\pi/4$, so this metric is incomplete. Every curve joining $(0,0)$ to $(0,L)$ has length at least $\int\cos u\,|v'|\,dt\geq |L|/\sqrt2$, because $\cos u\geq1/\sqrt2$ on the strip. Therefore $\boxed{\operatorname{diam}M=\infty}$ despite the positive Ricci bound. Completeness cannot be omitted.
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