Let $X$ be a topological space, $Y$ be a set and $f:X\to Y$ be a map of sets. We define the final topology $\mathscr{T}_f$ on $Y$ relative to $f$ to be the largest topology on $Y$ making $f$ continuous; this topology is exactely defined by $$\mathscr{T}_f=\{U\subseteq Y : f^{-1}(U) \ is \ open \ in \ X \}$$ Now consider an equivalence relation $\sim $ in $X$ and $\pi:X\to X/\sim$ the quotient map. The quotient topology in $X/\sim$ is defined to be the final topology on $X/\sim$ relative to $\pi$. Now if $X$ and $Y$ are topological spaces and $f:X\to Y$ is a surjection and a quotient map i.e, $U\subseteq Y$ is open iff $f^{-1}(U)\subseteq X$ is open. Then we define in $X$ the relation $\sim$ by: $x\sim x^\prime \Leftrightarrow f(x)=f(x^\prime)$. It is easy to see that $\sim$ is an equivalence relation and that $X/\sim$ endowed with the quotient topology is homeomorphic to $Y$.
Your "proof" simply uses the definition of continuity for the map $\pi$. That is, you've simply shown why $\pi$ is continuous, not its inverse. However, don't confuse the notation for $\pi^{-1}(U)$, which stands for the pre-image of $U$ under $p$, for the inverse of a function. The notation $\pi^{-1}$ one sees in the definition of continuity always means the pre-image - it's not, in general, a function. We are not necessarily assuming existence of inverses in the definition of contunity. In fact, we can even say that in general, the projection map does not have an inverse. As an example, we can define an equivalence relation $\sim$ on $\mathbb{R}^2$ by letting $(x,y) \sim (x',y')$ iff $y = y'$. The map $p: \mathbb{R}^2 \to \mathbb{R}^2 / \sim$ defined by $p(x,y) = [y]$ will not have an inverse $g: \mathbb{R}^2 / \sim \to \mathbb{R}^2$. For instance, what would $g([2])$ be? It could be $(1,2)$. But it could just as likely be $(0,2)$ or $(-100, 2)$. The problem is that the projection map $p$ is not necessarily injective.
Now, if we assume the projection map is bijective (technically only injectivity is necessary since the projection map is surjective, by definition), we can ask whether or not your characterization would be correct. In fact, the answer is yes. I'll state this formally:
Proposition: Let $\sim$ be an equivalence relation on a topological space $Q$, and suppose the projection map $p: Q \to Q / \sim$ given by $p(x) = [x]$ is injective. Then, the inverse of $p$, which we shall denote by $g: Q / \sim \to Q$, is continuous iff $p$ is open.
Proof: Let $g: Q / \sim \to Q$ be the inverse of the projection map $p: Q \to Q / \sim$, and assume $p$ is open. Now, let $U \subset Q$ be an open set. We must show $g^{-1}(U)$ is open. But thanks to the fact $p$ is bijective, we know that $g^{-1}(U) = p(U)$. As $p$ was an open map, by assumption, we conclude $p(U)$ is open. Thus, $g^{-1}(U)$ is open implying that the inverse function $g$ is continuous. To prove the converse, suppose that $g$ is continuous, and let $U \subset Q$ be open. By the bijectivity of $p$, we have $p(U) = g^{-1}(U)$, and as $g$ was continuous by assumption, $g^{-1}(U)$ is open $\implies p(U)$ is open. Thus, $p$ is an open map.
Best Answer
Yes, and this is just a matter of chasing the definitions. Suppose that $a\in \bar{f}^{-1}(U)$. Then $a=\nu_{\sim}(b)$ for some $b\in X$, and $\nu_{\sim'}(f(b))=\bar{f}(a)\in U$. Thus $b\in f^{-1}(\nu_{\sim'}^{-1}(U))$ and so $a\in \nu_{\sim}(f^{-1}(\nu_{\sim'}^{-1}(U)))$.
Conversely, suppose $a\in \nu_{\sim}(f^{-1}(\nu_{\sim'}^{-1}(U)))$. Then $a=\nu_{\sim}(b)$ for some $b\in f^{-1}(\nu_{\sim'}^{-1}(U))$. We then have $\bar{f}(a)=\nu_{\sim'}(f(b))\in U$ and thus $a\in \bar{f}^{-1}(U)$.
Note though that this does not solve the problem you started from in the way you describe, since contrary to what you say, $\nu_{\sim}$ is not typically an open map. To conclude that $\nu_{\sim}(f^{-1}(\nu_{\sim'}^{-1}(U)))$ is open you need to show not just that $f^{-1}(\nu_{\sim'}^{-1}(U))$ is open but also that it is saturated with respect to $\sim$ (i.e., if $x\in f^{-1}(\nu_{\sim'}^{-1}(U))$ and $y\sim x$ then $y\in f^{-1}(\nu_{\sim'}^{-1}(U))$).