Let $\psi\in H_A \otimes H_B$ be an entangled state, which means that it has Schmidt rank $r \geq 2$. Does there exist some $\epsilon>0$ for which all states $\varphi$ with $\|\psi - \varphi\|< \epsilon$ have Schmidt rank at least $r$? I think this must be studied somewhere but I'm unable to locate a reference or proof.
I assume $H_A$ and $H_B$ are finite dimensional.
TL;DR: Yes. The set of pure bipartite quantum states with Schmidt rank at least $r$ is open$^1$.
Let $S_{r}$ denote the set of all pure bipartite quantum states in $H_A\otimes H_B$ with Schmidt rank exactly $r$. Clearly, $H_A\otimes H_B=\bigcup_{i=1}^d S_i$ where $d=\min(\dim H_A, \dim H_B)$. For any $k\in\mathbb{N}$, define $S_{\leq k}=\bigcup_{i=1}^k S_i$ and $S_{\geq k}=\bigcup_{i=k}^d S_i$. I will show that all $S_{\leq k}$ are closed and all $S_{\geq k}$ are open.
Consider the map $s:H_A\otimes H_B\to\mathbb{R}^d$ which sends a pure bipartite quantum state to$^2$ the $d$-tuple of its Schmidt coefficients listed in decreasing order and padded$^3$ with zeros. Note that $s$ is continuous$^4$. Moreover, for any $k\in\mathbb{N}$ the set $\mathbb{R}^{(k)}=\mathbb{R}^k\times\{0\}^{d-k}$ is a closed subset of $\mathbb{R}^d$. But $S_{\leq k}=s^{-1}\big[\mathbb{R}^{(k)}\big]$. A continuous preimage of a closed set is itself closed, so $S_{\leq k}$ is closed. Therefore, its complement $$ \big(S_{\leq k}\big)^c=H_A\otimes H_B\setminus S_{\leq k}=S_{\geq k+1} $$ is open. Consequently, if $|\psi\rangle\in H_A\otimes H_B$ has Schmidt rank $r\geq 2$, then there is an $\epsilon>0$ such that every $|\phi\rangle$ with $\||\psi\rangle-|\phi\rangle\|<\epsilon$ has Schmidt rank at least $r$.
$^1$ In the usual topology induced by any of the norms equivalent to the operator norm.
$^2$ The image of $s$ is actually a subset of the unit sphere in $\mathbb{R}^d$.
$^3$ Padding is needed since Schmidt coefficients are positive real numbers by definition.
$^4$ The squares of the Schmidt coefficients are the eigenvalues of the reduced density matrix. Partial trace, eigenvalues and square root are continuous.