If $X = \{ |p(z)|I am trying problems in complex analysis of an institute in which I don't study and I got struck on this particular problem.

Let $p$ be a non-constant polynomial, $c>0$ and $X=\{z:p(z)<c\}$. Prove that $\partial X=\{z:|p(z)|=c\}$ and also each connected component of $X$ contains a zero $p$.

Attempt : Let $C$ be connected component of $X$ . Then $\partial C$ is a  subset of $\partial X$ and so $|p(z)| \leq c$. But I cant think any other argument other than this.
So, kindly shed some light on how this problem should be approached.
 A: Simple properties of $p$:
i) Since $p$ is a nonconstant polynomial, $p$ takes on every complex value. Thus the sets $X=\{|p|<c\}$ and $\{|p|=c\}$ are nonempty.
ii) The set $X$ is open by the continuity of $|p|.$
iii) $|p(z)|\to \infty$ as $|z|\to\infty.$
From iii) it follows that $X$ is bounded. Otherwise $X$ would contain a sequence $z_n$ such that $|z_n|\to \infty,$ hence $|p(z_n)|\to \infty,$ violating the definition of $X.$
Let $z\in \partial X.$ Then $z$ is the limit of a sequence in $X.$ This implies $|p(z)|\le c.$ Could $|p(z)|<c$ happen? No, because then $z\in X$ and it couldn't be a boundary point. It follows that $\partial X\subset \{|p|=c\}.$
Now suppose $|p(z)|=c.$ Let $r>0.$ Then $p(D(z,r))$ is open by the open mapping theorem, hence contains points of modulus less than $c$ and points of modulus greater than $c.$ Thus $D(z,r)\cap X$ and $D(z,r)\cap X^c$ are both nonempty. Since $r$ was arbitrary, $z\in \partial X.$ This with the last paragraph proves $\partial X = \{|p|=c\}.$
Recall the maximum modulus theorem: Suppose $U$ is a bounded open connected set. Let $f$ be continuous on $\overline U$ and holomorphic on $U.$ If the maximum of $|f|$ occurs in $U,$ then $f$ is constant.
So now let $C$ be a connected component of $X.$ We know $\partial C \subset \partial X,$ which implies $|p|=c$ on $\partial C.$ And of course $|p|<c$ in $C.$
Assume $C$ does not contain a zero of $p.$ Then $p$ is nonzero on $\overline C,$ a compact set. Thus $|p|$ attains a positive minimum at some in $z_0 \in \overline C.$ By the MMT, $z_0\in C.$ But notice $1/p$ satisfies the hypotheses of the MMT. Thus, using MMT again,
$$\frac{1}{|p(z_0)|} < \max_{\partial C}\frac{1}{|p|} =\frac{1}{c}.$$
Since $|p(z_0|<c,$ we have a contradiction, and we're done.
