If only weak-ordering and continuity is assumed, "ICs" can definitely intersect.
If we assume Monotonicity or convexity in addition to weak-ordering, then we can get "no cross of IC".
But those two assumptions are argueably too strong. In some real world scenarios they are violated.
Are there any weaker assumptions for the "IC cannot cross"?
If only weak-ordering and continuity is assumed, ICs can definitely intersect.
This is not true. First, if you're speaking of indifference curves, you'd already be assuming either local non-satiation or monotonicity.
Let's speak of indifference sets instead. The analogue of two sets, $I_1$ and $I_2$, "crossing" each other can be formalized as $I_1\ne I_2$ and $I_1\cap I_2\ne \varnothing$.
Take two alternatives $x_1,x_2\in X$ and define two indifference sets as follows: \begin{align} I_1:=\{x\in X:x\sim x_1\}\quad\text{and}\quad I_2:=\{x\in X:x\sim x_2\}. \end{align} WLOG, assume that $x_1\succ x_2$, so that $I_1\ne I_2$. If we allow $I_1$ to "cross" $I_2$, then $I_1\cap I_2$ must be non-empty. Let $\bar x\in I_1\cap I_2$ be an element in the intersection. By the transitivity of the indifference relation $\sim$, we have $x_1\sim \bar x$ and $\bar x\sim x_2$, implying $x_1\sim x_2$. But this is contradictory to our assumption that $x_1\succ x_2$. The contradiction therefore suggests that transitivity, a property of the weak ordering, is violated.
Consider any utility function $u: \mathbb{R}^2_+ \rightarrow\mathbb{R}$. Indifference curve for satisfaction level $\mu$ is defined as: $\text{IC}(\mu) = \{(x, y)\in\mathbb{R}^2_+|u(x, y) = \mu\}$
I am assuming that by ICs cannot cross, you mean that ICs for two different utility levels don't have a non-empty intersection i.e.
$(\forall (\mu_1, \mu_2)\in\mathbb{R}^2) (\mu_1 \neq \mu_2 \Rightarrow \text{IC}(\mu_1) \cap \text{IC}(\mu_2) = \emptyset)$.
This is obviously true because of the fact that $u$ is a function.
Related Question: Can an IC look like a cross? https://qr.ae/pG0AoL
