As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi_n(x)=0\}$ is empty. The measure of interior zeros is neither continuous: if $\psi_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi_n(x)=0\}$ is a circle of radius $1-o(1).$

You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi_n$ by $N\ge0$ closed curves $\{\gamma_{j,n}:\mathbb{S}^1\to\Omega\}_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$ and $\phi_n$, which is finite and definitively constant wrto $n$. Then the length of each component $\int_{\mathbb{S}^1}|\dot\gamma_{j,n}(s)|^2ds$ passes to the limit.

As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi_n(x)=0\}$ is empty. The measure of interior zeros is neither continuous: if $\psi_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi_n(x)=0\}$ is a circle of radius $1-o(1).$

You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi_n$ by $N\ge0$ closed curves $\{\gamma_{j,n}:\mathbb{S}^1\to\Omega\}_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$ and $\phi_n$, which is finite and definitively constant wrto $n$. Then the length of each component $\int_{\mathbb{S}^1}|\dot\gamma_{j,n}(s)|ds$ passes to the limit.

As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi_n(x)=0\}$ is empty. The measure of interior zeros is neither continuous: if $\psi_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi_n(x)=0\}$ is a circle of radius $1-o(1).$

You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi_n$ by $N\ge0$ closed curves $\{\gamma_{j,n}:\mathbb{S}^1\to\Omega\}_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$, which is finite and definitively constant wrto $n$.

As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi_n(x)=0\}$ is empty. The measure of interior zeros is neither continuous: if $\psi_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi_n(x)=0\}$ is a circle of radius $1-o(1).$

You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi_n$ by $N\ge0$ closed curves $\{\gamma_{j,n}:\mathbb{S}^1\to\Omega\}_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$ and $\phi_n$, which is finite and definitively constant wrto $n$. Then the length of each component $\int_{\mathbb{S}^1}|\dot\gamma_{j,n}(s)|^2ds$ passes to the limit.