Stability of the nonempty intersection of an open set $A$ with a set $S$ under homotopy? To be more precise: let $F(x,t) : R^2 \times I \to R^2$ be a homotopy of open maps $F(_,t)(x)$ (the restriction of $F$ to some fixed $t$) (the homotopy is continuous in both variables). Suppose that for some open set $V \subset R^2$ and arbitrary set $S \subset R^2$, $F(V,0) \cap S \not = \emptyset$. Is it the case that for sufficiently small $t$, $F(V,t) \cap S \not = \emptyset$?
We can assume that $S = \{s\}$, and that $F(a,0) = s$. 
I have a few ideas about how to proceed, though none of them have worked yet. However, I don't want to spend that much more time arguing for something false. (This lemma would be helpful for a complex analysis exercise, but I probably don't need this level of generality.)
Is this proposition true? (What about if I insist that each $F(,t)(x)$ is holomorphic, in addition to being open?)
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I have a sketch of an argument in the case when $F$ is a linear homotopy between two open sets $U_0$ and $U_1$ in $\mathbb{C}$.
We assume $s = 0$ for notational simplicity.
As $0 \in U$, there is a loop $\gamma$ around $s$ in $U$. The homotopy from $U_0$ to $U_1$ restricts to a homotopy of $\gamma$, which we denote $\gamma_t$. Now let $G(t) = \int_{\gamma_t} 1/z dz$, which is continuous in $t$ since $1/z$ is continuous. $G(t) = 2\pi i$ if $t = 0$, and $G(t) = 0$ if $t = 1$, as the pole is or is not inside the loop. Since $G$ is continuous in $t$, there is thus some $\epsilon > 0$ so that $\forall t < \epsilon$, $G(t) = 2\pi i$. This implies that for all of those $t$, $0$ is in the interior of the loop $\gamma_t$.
Now, since $\pi_1$ is a homotopy invariant, the image of the disc of $\gamma_0$ under this homotopy, $D_t$, is simply connected. Since $D_t$ contains $\gamma_t$ the disc of $\gamma_t$ must be filled in: $DiscOf(\gamma_t) \subseteq D_t \subseteq U_t$. Therefore, since $0 \in D_t$, $0 \in U_t$ for $t < \epsilon$, which is the claim.
An issue with my argument: it is not completely clear that the $\gamma_t$ remain simple curves under this homotopy, even for small $t$.
However, I can allow myself to assume that this linear homotopy between open sets is actually the result of a homotopy between functions $f$ and $g$ that are holomorphic on the open domain $V$. 
If I can guarantee the existence of a path in $U_0 = f(V)$ around 0 that comes from a path in $V$, then I believe the rest of the proof goes through. This seems intuitive, but I'm not sure how to construct it. I'll ask another question about that, because this one is starting to get long winded.
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All of this seems to come down to this question:
Given a holomorphic $f: V \to W$, with $V$ and $W$ open sets, is there a loop around $f(a)$ in $W$ that is the image under $f$ of a smooth loop around a? 
I'm pretty sure that the answer to this is affirmative - in that the local picture of a holomorphic map is of the form $z^k$ up to homeomorphism.
 A: Clearing up a terminological uncertainty first: A homotopy is, by definition, continuous in both variables.
Now to the problem.
To simplify notation, let's say $s = 0$, and $a = 0$.
If we assume $f = F(\,\cdot\,,0)$ is holomorphic, we can show the result. We need not the full consequences of holomorphicity, only the existence of a certain curve. Namely, since $f$ is holomorphic and non-constant, $0$ is an isolated zero of $f$, so there is a $\rho > 0$ such that $\overline{D_\rho(0)} \subset V$ such that $f$ has no zeros on the circle $\{z : \lvert z\rvert = \rho\}$. Then
$$d := \frac{1}{2\pi i} \int_{\lvert z\rvert = \rho} \frac{f'(z)}{f(z)}\,dz$$
is the number of zeros of $f$ in $D_\rho(0)$, counting multiplicity, so $d > 0$ since $f(0) = 0$. On the other hand, $d$ is the winding number of the curve $s \mapsto f(\rho e^{is}),\, s\in [0,2\pi]$.
That is the curve we need, a simple closed Jordan curve $\gamma$ in $V$ with $n(\gamma,0) = 1$ and $n(f\circ \gamma, 0) \neq 0$.
If we have such a curve, we choose an $\varepsilon > 0$ small enough that $F(\gamma(s),t)\neq 0$ for all $s\in [0,2\pi]$ and $t\in [0,\varepsilon]$, and consider $h \colon S^1\times[0,\varepsilon] \to \mathbb{R}^2\setminus \{0\}$ given by
$$h(e^{is},t) = F(\gamma(s),t).$$
Then $h$ is a homotopy of closed curves in $\mathbb{R}^2\setminus\{0\}$, so all $\alpha_t \colon e^{is} \mapsto F(\gamma(s),t)$ have the same degree, in particular $\deg \alpha_t \neq 0$ for all $t\in [0,\varepsilon]$. That means that no $\alpha_t$ has a continuous extension to a map $D^2 \to \mathbb{R}^2\setminus\{0\}$, since such an extension would provide a homotopy to a constant map, and that would imply $\deg\alpha_t = 0$. But $F(\,\cdot\,,t)$ provides an extension of the curve $\alpha_t$ to a continuous map $D^2\to\mathbb{R}^2$, which implies that $F(\,\cdot\,,t)$ has a zero in the interior of $\gamma$, which is contained in $V$.
