**Yes.**

With an appropriate topology, the function mapping a non-zero harmonic function to its zero set is continuous. The result actually applies to open mappings from a locally compact locally connected space $X$ to $\mathbb{R}$, and harmonic functions are just a particular example of such open mappings.

If $X$ is a topological space, then let $H(X)$ denote the collection of all closed subsets of $X$. The set $H(X)$

If $X$ is a compact Hausdorff space and $U,V_1,\dots,V_n$ are non-empty open subsets of $X$, then let $[U;V_1,\dots,V_n]$ be the collection of all closed subsets $C\subseteq X$ where $C\subseteq U$ and where $C\cap V_1\neq\emptyset,\dots,C\cap V_n\neq\emptyset$. Then the Vietoris topology on $H(X)$ is the topology with basis consisting of all sets of the form $[U;V_1,\dots,V_n]$.

Proposition: Let $X$ be a locally compact locally connected Hausdorff space. Let $H$ be a collection of continuous open mappings $f:X\rightarrow\mathbb{R}$. Give $H$ the topology of uniform convergence on compact sets. Define a mapping $T:H\rightarrow X\cup\{\infty\}$ by letting $T(h)=Z(h)\cup\{\infty\}$. Then the function $T$ is continuous.

Proof: Suppose that $h\in H$. Let $z\in Z(h)$. Let $U$ be a neighborhood of $z$. Then there is some connected open set $V$ with $z\in V\subseteq\overline{V}\subseteq U$ and where $V$ is compact. In this case, we have $h(r)>0,h(s)<0$ for some $r,s\in V$. Let $\delta=\min(h(r),-h(s))$. Therefore, if $g\in H$ and $\|g|_U-h|_U\|<\delta$, then $g(r)>0,g(s)<0$. Since $V$ is connected and $r,s\in V,g(r)>0,g(s)<0$, there is some $t\in V$ with $g(t)=0$ (this is the reasoning that Giorgio Metafune has made in the comments).

We conclude that if $Z(h)\cap U\neq\emptyset$, then there is some $\delta>0$ where if $\|(g-h)|_U\|<\delta$, then $Z(g)\cap U\neq\emptyset$ as well.

Suppose now that $h\in H$, and $T(h)\in[U;V_1,\dots,V_n]$ for open subsets $U,V_1,\dots,V_n\subseteq X\cup\{\infty\}$. Then $Z(h)\cup\{\infty\}\subseteq U$ and $Z(h)\cup\{\infty\}\cap V_j\neq\emptyset$ for $1\leq j\leq n$. Without loss of generality, assume that $1\leq m\leq n$ and $\infty\not\in V_j$ for $1\leq j\leq m$ where $\infty\in V_j$ for $m<j\leq n$. Then $V_j$ is an open subset of $X$ with $Z(h)\cap V_j\neq\emptyset$ for $1\leq j\leq m$.

Therefore, for $1\leq j\leq m$ there is a relatively compact open $W_j\subseteq V_j$ and some $\delta_j$ where if $\|(g-h)|_{W_j}\|<\delta_j$, then $Z(g)\cap V_j\neq\emptyset$. Now, set $W=W_1\cup\dots\cup W_m$ and $\delta=\min(\delta_1,\dots,\delta_m)$. Then whenever $\|(g-h)|_W\|<\delta$, we have $Z(g)\cap V_j\neq\emptyset$ for $1\leq j\leq m$.

Now let $C=X\setminus U$. Then $C$ is a compact subset of $X$ with $0\not\in g[C]$. Therefore, there is some $\epsilon>0$ where $|g(c)|\geq\epsilon$ for each $c\in C$. Therefore, if $\|(h-g)|_C\|<\epsilon$, then $g(c)\neq 0$ whenever $c\in C$, so $Z(g)\subseteq U$.

Therefore if $g\in H$ and $\|(h-g)|_{C\cup W}\|<\min(\delta,\epsilon)$, then
$g\in[U;V_1,\dots,V_n]$ as well. We therefore conclude that the function $T$ is continuous at the point $h$.

Q.E.D.