I am confused due to the following lines of reasoning:
$(1)$ $K$ is compact and $g_n$ is continuous and therefore bounded and attains its maximum, say $M$, somewhere on $K$. The interval $[\varepsilon,M]$ is closed and so its inverse must be closed.
$(2)$ Let $\varepsilon=g(x_0)-a$. Since $g_n$ is continuous we can choose an $s <a$ such that $|g_n(x)-g_n(x_0)|<s$ whenever $x$ is in a neighborhood of radius $\delta$ of $x_0$. The values of $g_n(x)$ at points in the neighborhood are either bigger than $g_n(x_0)$ and so bigger than $\varepsilon$ or $g_n(x)>g_n(x_0)-s = \varepsilon + a-s> \varepsilon$ so that all points of the neighborhood are in $K_n$ and $x_0$ is an interior point of $K_n$ which shows that $K_n$ is open.
Note that if $g_n(x_0) = \varepsilon$ then $a=0$, so you haven't shown that every point in $K_n$ is an interior point. I think $(1)$ is correct.