2.  Suppose furthermore that $X\subset{{\mathbb R}}^2$, that $U\subset X$ is an open set on which the functions $f, h_t$ are harmonic, and that $f$ is nowhere locally identically zero on $U$. Then, for all $\epsilon > 0$, there exists $t_0=t_0(\epsilon)>0$ such that $Z \cap U \subset N_{\epsilon}(Y_t \cap U)$ for all $t<t_0$.

Suppose that (1) fails: that is, there exist $\epsilon > 0$ and sequences $t_n
\rightarrow 0$, $y_n \in Y_{t_n}$ such that for all $n$ we have $B(y_n, \epsilon) \cap Z =
\emptyset$. Since $X$ is compact, the sequence $\{y_n\}$ has a limit point; therefore, replacing $y_n$ and $t_n$ by suitable subsequences, we may assume without loss of generality that the sequence $y_n$ converges to a point $y$.

Note that $y \not\in Z$. Indeed, $B(y,\epsilon/2) \subset B(y_n,\epsilon)$ for $n$ sufficiently large, so in fact $B(y,\epsilon/2) \cap Z = \emptyset$. Choose a connected neighborhood $U \subset B(y,\epsilon/2)$ of $y$; by the Intermediate Value Theorem, $f$ is either strictly positive or strictly negative on $U$. Since $f$ is continuous and $X$ is compact, we find a neighborhood $V \subset U$ of $y$ and a constant $\delta > 0$ such that $$\label{eq:delta}
    |f(x)| > \delta > 0 \qquad \text{for all} \ x 