Let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$, $X_G:=\{H\subseteq G\mid H\le G \wedge H\ne G\}$. Lagrange's theorem put a limitation on the order of the elements of $X_G$, which must be a proper divisor of $|G|$. If $G$ is simple, then an additional limitation on the order of the elements of $X_G$ gets in, namely:
$$G\space\text{simple}\Longrightarrow[G:H]!\ge|G|, \space\forall H\in X_G \tag 1$$
In fact, for $H\in X_G$, the group $G$ acts by left multiplication on the left quotient set $G/H$, and this action has trivial kernel$^\dagger$. So, $G$ embeds into $S_{[G:H]}$, whence $(1)$.
The constraint $(1)$ means that finite simple groups can't have "relatively big" subgroups. For example, from $(1)$ follows that $A_5$ can't have subgroups of order $15$, $20$ and $30$ (this latter is ruled out by the very simplicity of $A_5$, though), because $[60:k]!<60$ for $k=15, 20, 30$.
Does the inverse implication in $(1)$ hold, too? Some thoughts. A counterexample would be a nonsimple finite group with all its proper subgroups fulfilling $(1)$. $|G|=2,3$: it is $G\cong C_2,C_3$, both simple, so we have to move upwards. $|G|=4$: both classes ($C_2\times C_2$ and $C_4$) are nonsimple, but both have subgroups of index $2$ (and $2!<4$). $|G|=5$: $C_5$ is simple. $|G|=6$: likewise $|G|=4$ case, both classes ($S_3$ and $C_6$) are nonsimple, but both have subgroups of index $2$ (and $2!<6$). $|G|=7$: $C_7$ is simple. Therefore, counterexamples (if any) must have $|G|\ge 8$.
$^\dagger$For $H\in X$, we get: $K:=\bigcap_{g\in G}gHg^{-1}\lneq G$, and thence $K=\{1\}$ for the simplicity of $G$.
Best Answer
Let $p$ be a prime and $k > 1$ such that $p^k \leq p!$. Then any finite group $G$ of order $p^k$ satisfies your condition, since any subgroup of $G$ has index $\geq p$. (For example consider $G$ elementary abelian of order $25 = 5^2$.)
You could also look any group of order $n = p_1^{k_1} \cdots p_t^{k_t}$, where $p_i$ are distinct primes such that $n \leq p_i!$ for all $1 \leq i \leq t$.