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In general, if we do not assume that $f$ is proper (I missed this assumption when I read the question), $c$ need not be a regular value of $g$ for any $\epsilon>0$. For example $0$ is a regular value of $f(x)=e^x\sin x$, but it approaches to $0$ as $x\to -\infty$ and you can find arbitrarily small perturbations $g$ of $f$ so that $g^{-1}(0)$ will contain an interval. For example if $\phi\in C^\infty(\mathbb{R})$, $\phi(x)=0$ for $|x|\leq 1$ and $\phi(x)=1$ for $|x|\geq 2$, then $g_m(x)=\phi(x+m)e^x\sin x$ will converge to $f(x)=e^x\sin x$ in $C^1$ topology as $m\to\infty$, but $g_m^{-1}(0)$ will contain the interval $[-m-1,-m+1]$.

In general, if we do not assume that $f$ is proper (I missed the word "proper" when I read the question), $c$ need not be a regular value of $g$ for any $\epsilon>0$. For example $0$ is a regular value of $f(x)=e^x\sin x$, but it approaches to $0$ as $x\to -\infty$ and you can find arbitrarily small perturbations $g$ of $f$ so that $g^{-1}(0)$ will contain an interval. For example if $\phi\in C^\infty(\mathbb{R})$, $\phi(x)=0$ for $|x|\leq 1$ and $\phi(x)=1$ for $|x|\geq 2$, then $g_m(x)=\phi(x+m)e^x\sin x$ will converge to $f(x)=e^x\sin x$ in $C^1$ topology as $m\to\infty$, but $g_m^{-1}(0)$ will contain the interval $[-m-1,-m+1]$.

In general, $c$ need not be a regular value of $g$ for any $\epsilon>0$. For example $0$ is a regular value of $f(x)=e^x\sin x$, but it approaches to $0$ as $x\to -\infty$ and you can find arbitrarily small perturbations $g$ of $f$ so that $g^{-1}(0)$ will contain an interval. For example if $\phi\in C^\infty(\mathbb{R})$, $\phi(x)=0$ for $|x|\leq 1$ and $\phi(x)=1$ for $|x|\geq 2$, then $g_m(x)=\phi(x+m)e^x\sin x$ will converge to $f(x)=e^x\sin x$ in $C^1$ topology as $m\to\infty$, but $g_m^{-1}(0)$ will contain the interval $[-m-1,-m+1]$.

In general, if we do not assume that $f$ is proper (I missed this assumption when I read the question), $c$ need not be a regular value of $g$ for any $\epsilon>0$. For example $0$ is a regular value of $f(x)=e^x\sin x$, but it approaches to $0$ as $x\to -\infty$ and you can find arbitrarily small perturbations $g$ of $f$ so that $g^{-1}(0)$ will contain an interval. For example if $\phi\in C^\infty(\mathbb{R})$, $\phi(x)=0$ for $|x|\leq 1$ and $\phi(x)=1$ for $|x|\geq 2$, then $g_m(x)=\phi(x+m)e^x\sin x$ will converge to $f(x)=e^x\sin x$ in $C^1$ topology as $m\to\infty$, but $g_m^{-1}(0)$ will contain the interval $[-m-1,-m+1]$.

In general, $c$ need not be a regular value of $g$ for any $\epsilon>0$. For example $0$ is a regular value of $f(x)=e^x\sin x$, but it approaches to $0$ as $x\to -\infty$ and you can find arbitrarily small perturbations $g$ of $f$ so that $g^{-1}(0)$ will contain an interval.

PS. I edited your question, because contrary to what you said, $f^{-1}(c)$ need not be compact. Please, think about your question more carefully and ask it as a new one instead of editing this one.

In general, $c$ need not be a regular value of $g$ for any $\epsilon>0$. For example $0$ is a regular value of $f(x)=e^x\sin x$, but it approaches to $0$ as $x\to -\infty$ and you can find arbitrarily small perturbations $g$ of $f$ so that $g^{-1}(0)$ will contain an interval. For example if $\phi\in C^\infty(\mathbb{R})$, $\phi(x)=0$ for $|x|\leq 1$ and $\phi(x)=1$ for $|x|\geq 2$, then $g_m(x)=\phi(x+m)e^x\sin x$ will converge to $f(x)=e^x\sin x$ in $C^1$ topology as $m\to\infty$, but $g_m^{-1}(0)$ will contain the interval $[-m-1,-m+1]$.