I'm working independently through Wallace's "An Introduction to Algebraic Topology" and I'm unsure of a claim in one of the exercises. The question states
II.2.2. Let $M_n$ denote the set of square $n$-rowed matrices with real elements. If the elements are taken as coordinates in $n^2$-dimensional Euclidean space $M_n$ becomes a topological space. Prove that the set of matrices of rank greater than or equal to $r$ for any $r\le n$ is an open set in this space.
My understanding of the question is that the Euclidean space on $\mathbb R^{n^2}$ gives $M_n$ a topology in the sense that a set $U_A$ of $A\in M_n$ is a neighborhood of $A$ if there is a bijective $f:M_n\to \mathbb R^{n^2}$ such that the image $f(U_A)$ is a neighborhood of $f(A)$.
I have no problems with $M_n$ being a topological space because if $A,B\in M_n$ we can define $d(A,B)=\|A-B\|_F$ with $\|\cdot\|_F$ as the Frobenius norm and it's clear then that if $\forall\alpha,\beta\in\mathbb R^{n^2}$,$d(\alpha,\beta)$ is the Euclidean distance, then $d(A,B)=d(f(A),f(B))$. I suspect that he threw in the reference to $n^2$-Euclidean space to bypass the Frobenius norm.
Anyway, I've come up with the following counterexample to the claim that the set of matrices of rank greater to or equal to $r$ for any $r\le n$ is an open set in the space:
We can define $f:M_4\to\mathbb R^{16}$ with $$f(A)=(a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10},a_{11},a_{12},a_{13},a_{14},a_{15},a_{16})$$ if $A\in M_4$ with $$A=\begin{bmatrix}a_1&a_2&a_3&a_4\\a_5&a_6&a_7&a_8\\a_9&a_{10}&a_{11}&a_{12}\\a_{13}&a_{14}&a_{15}&a_{16}\end{bmatrix}.$$ Let $\epsilon \gt 0$, $a\in\mathbb R$, and $b\lt a+\frac{\epsilon}{2}$. Then, define $A,B\in M_4$ by $$A= \begin{bmatrix}0&0&0&0\\ 0&0&0&0\\ a&b&a&b\\ b&a&b&a\end{bmatrix} \text{ and } B=\begin{bmatrix}0&0&0&0\\ 0&0&0&0\\ a&b&a&b\\ a&b&a&b\end{bmatrix}$$
Then, \begin{align}d(f(A),f(B))&=\sqrt{4(b-a)^2}\\&=2(b-a)\\&\lt2(a+\frac{\epsilon}{2}-a)\\&=\epsilon\end{align} So, rank $A=2$ and rank $B=1$. Define $O=\{ M\in M_4: \text{rank } M\ge 2\}$. Clearly, $A\in O$, but $B\notin O$. However, $f(B)$ is in every $\epsilon$-neighborhood of $f(A)$. So, $B$ must be in every neighborhood of $A$, so $O$ is not a neighborhood of $M_4$. Since $O$ is not a neighborhood of $M_4$, it cannot contain any neighborhoods of $A$, so $O$ cannot be open.
I was unable to locate any errata for this book, so there may have been a typo, but it's more likely that my understanding of the question is incorrect. So, assuming that the exercise's assertion is correct, where is the flaw in my counterexample?
But your $A,B$ are really moving targets, because $b$ depends on $\epsilon$ and hence so do the corresponding matrices. Really, they should be denoted $A_\epsilon, B_\epsilon.$ So you've shown $d(f(A_\epsilon), f(B_\epsilon)) < \epsilon,$ but that's no contradiction.