I know that the zero-set of a non-zero polynomial in $\mathbb{C}[x\_1,...,x\_n]$ can not have interior points, but I'm trying to find a proof that doesn't require a knowledge of complex analysis like holomorphic functions, etc...

So, suppose that $V=\{(x\_1,\cdots, x\_n)\in \mathbb{C}^n: p(x\_1,\cdots,x\_n)=0\}$. Now if there is an interior point in $\mathbf{y}=(y\_1,\cdots,y\_n) \in V$ then there exists $\epsilon>0$ such that $\displaystyle N\_\epsilon(\mathbf{y}) \subseteq V$. But we know that in every metric space open balls are convex, so for all $\lambda \in \mathbb{R}$ the point $\mathbf{\bar{y}}=\lambda\cdot\mathbf{y}+(1-\lambda)\cdot \mathbf{y'} \in N\_\epsilon(\mathbf{y})$ lies in $V$where $\mathbf{y'} \in N\_\epsilon(\mathbf{y})$.

If I plug $\mathbf{\bar{y}}$ into $p(x\_1, \cdots, x\_n)=0$ I'll get a new non-zero polynomial $p\_1(x\_1,\cdots,x\_n,\lambda)=0$. Now, if I fix $x\_1,\cdots,x\_n$ and consider $p\_1(x\_1,\cdots,x\_n,\lambda)$ as a polynomial only in $\lambda$ then I see that $f(\lambda)=p\_1(x\_1,\cdots,x\_n,\lambda)$ then because of convexity $f(\lambda)$ must vanish on every $\lambda \in [0,1]$, but from the fundamental theorem of algebra we know that every non-zero polynomial in $\mathbb{C}[\lambda]$ can have a finite number of roots, so, $f(\lambda)=0$. But I don't see how $f(\lambda)=0$ could lead me to a contradiction.