Myers' theorem

By the Hopf–Rinow theorem, since $M$ is complete any two points are joined by a minimizing geodesic. Suppose toward a contradiction that some unit-speed minimizing geodesic $\gamma:[0,L]\to M$ has length $L > \pi/\sqrt{k}$.

Because $\gamma$ minimizes length, the second variation (index form) satisfies $I(V,V) \geq 0$ for every piecewise-smooth vector field $V$ along $\gamma$ vanishing at both endpoints, where $I(V,V) = \int_0^L \left(|V'|^2 - \langle R(V,\gamma')\gamma', V\rangle\right)dt$. Choose a parallel orthonormal frame $E_1,\dots,E_{n-1}$ along $\gamma$, everywhere orthogonal to $\gamma'$, and test with the fields $V_i(t) = \sin(\pi t/L)\,E_i(t)$ for $i=1,\dots,n-1$.

Since $E_i$ is parallel, $V_i' = \frac{\pi}{L}\cos(\pi t/L)E_i$, so $I(V_i,V_i) = \int_0^L\left[\left(\frac{\pi}{L}\right)^2\cos^2(\pi t/L) - \sin^2(\pi t/L)\langle R(E_i,\gamma')\gamma',E_i\rangle\right]dt$. Summing over $i$ and using $\sum_i \langle R(E_i,\gamma')\gamma',E_i\rangle = \mathrm{Ric}(\gamma',\gamma') \geq (n-1)k$ gives $\sum_i I(V_i,V_i) \leq (n-1)\int_0^L\left[\left(\frac{\pi}{L}\right)^2\cos^2(\pi t/L) - k\sin^2(\pi t/L)\right]dt$.

Since $\int_0^L\cos^2(\pi t/L)\,dt = \int_0^L\sin^2(\pi t/L)\,dt = L/2$, the right side equals $(n-1)\cdot\frac{L}{2}\left[\left(\frac{\pi}{L}\right)^2 - k\right]$, which is strictly negative whenever $L > \pi/\sqrt{k}$. So some $I(V_i,V_i) < 0$, contradicting that $\gamma$ minimizes length. Hence every minimizing geodesic satisfies $L \leq \pi/\sqrt{k}$, so $\mathrm{diam}(M) \leq \dfrac{\pi}{\sqrt{k}}$; by Hopf–Rinow, a complete manifold of finite diameter is compact, and the same argument applied to the universal cover (which inherits the same Ricci lower bound) shows it too is compact, forcing $\pi_1(M)$ to be finite.