Let $g_1,\cdots,g_m \in \mathbb{R}[x]=\mathbb{R}[x_1,\cdots,x_d]$ be polynomials with real coefficients and define the semi-algebraic sets $K=\left\{x \in \mathbb{R}^d: g_i(x) \ge 0, \forall i\right\}, K'=\left\{x \in \mathbb{R}^d: g_i(x) >0, \forall i\right\}$.
Proposition: The interior of $K$ is nonempty if and only if $K'$ is nonempty.
Proof: If $K' \neq \emptyset \Rightarrow Int(K) \neq \emptyset$, since $K' \subset Int(K)$. Conversely, suppose $Int(K) \neq \emptyset$ but $K'= \emptyset$. Then there exists an $\epsilon$-ball $\mathcal{B}$ inside $K$. Consider the polynomial $g=g_1 \cdots g_m$. Since $K'=\emptyset$, $g$ vanishes on $K$ and thus $g$ vanishes on the ball $\mathcal{B}$.
Question: Why does the vanishing of $g$ on $\mathcal{B}$ imply that $g=0$?
You can prove that if $g \in \mathbb R[x_1, \ldots, x_n]$ is zero on any epsilon ball then $g = 0$. You do this by induction on $n$. For the base case $n = 1$ use that non-zero polynomials only have finitely many roots. For the inductive step write $g = \sum_ix_n^ig_i$ where each $g_i$ is a polynomial in $x_1, \ldots, x_{n-1}$. If $g \neq 0$ then some $g_i \neq 0$ so there is a point $a = (a_1, \ldots, a_{n-1}) \in \mathbb R^{n-1}$ such that $g_i(a) \neq 0$ (you can choose $a$ such that for some $a_n$ the point $(a_1, \ldots, a_n) \in \mathbb R^n$ is in the epsilon ball). Then $g(a_1, \ldots, a_{n-1}, x_n)$ is a nonzero polynomial in one variable that vanishes on some epsilon ball, which is a contradiction.