= Solution
\b[False with the $\limsup$ printed in the paper.] Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Then
$$
\limsup_{n\to\infty}\inf_{x\in S_n}K_n(x)=1>0,
$$
but infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their <injectivity radii> have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|\partial_rK|$. Thus this alternating sequence satisfies all four additional conditions as well.
The intended statement becomes true if $\limsup$ is replaced by
$$
\liminf_{n\to\infty}\inf_{x\in S_n}K_n(x)\geq0.
$$
Here is the proof. Put $\delta_n=\max(0,-\inf_{S_n}K_n)$, so $\delta_n\to0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The <supporting-sphere curvature bound> gives
$$
K_n(p_n)\geq\kappa:=R^{-2}>0.
$$
Write $K_0=K_n(p_n)$ and $j(r,\theta)=\sqrt{G(r,\theta)}$ in <geodesic polar coordinates>. Choose a fixed sufficiently small $c>0$ and set $\rho_n=cK_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that $\rho_n\leq\epsilon_0$ and
$$
\frac12K_0\leq K_n(r,\theta)\leq\frac32K_0\qquad(0\leq r\leq\rho_n).
$$
The <Jacobi equation> $j_{rr}=-K_nj$, with $j(0)=0$ and $j_r(0)=1$, and the <Sturm comparison theorem> then give $j(r,\theta)\geq c_1r$ on this ball for a constant $c_1>0$ independent of $n$. Consequently
$$
\int_{B(p_n,\rho_n)}K_n\,dA
\geq\int_0^{2\pi}\int_0^{\rho_n}\frac{K_0}{2}c_1r\,dr\,d\theta
=\frac{\pi c_1c^2}{2}=:c_0>0.
$$
Since $K_n\geq-\delta_n$ elsewhere and $\operatorname{Area}(S_n)\leq M$,
$$
\int_{S_n}K_n\,dA\geq c_0-M\delta_n>0
$$
for all sufficiently large $n$. <Gauss-Bonnet theorem>[Gauss-Bonnet] forces $\chi(S_n)>0$, and the <classification theorem for surfaces> makes each such $S_n$ a sphere.
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