Let $S$ be an algebraic subset of $\mathbb{R}^d$ with vanishing ideal $I_S=(f_1(x),\cdots,f_m(x))$. Suppose that $S$ has infinite cardinality (countable or uncountable, does not matter). I want to prove that the dimension of the real vector space $\mathbb{R}[x_1,\cdots,x_d]/I_S$ is infinite.
Here is my attempt: We have that $\operatorname{dim}_{\mathbb{R}} \mathbb{R}[x_1,\cdots,x_d]/I_S \ge \operatorname{dim}_{\mathbb{C}} \mathbb{C}[x_1,\cdots,x_d]/I_S$, since if we view a real vector space as a complex vector space, its dimension can only drop. Now suppose that $\operatorname{dim}_{\mathbb{C}} \mathbb{C}[x_1,\cdots,x_d]/I_S$ is finite and equal to $n>0$. Then by Groebner basis theory, the complex variety defined by $I_S$ must have finite cardinality and less or equal to $n$. But we already know that a subset of that variety is $S$, which is infinite. Hence $\operatorname{dim}_{\mathbb{C}} \mathbb{C}[x_1,\cdots,x_d]/I_S =\infty$ and by the previous inequality on the dimensions we also have that $\operatorname{dim}_{\mathbb{R}} \mathbb{R}[x_1,\cdots,x_d]/I_S=\infty$.
Question: is my argument technically sound? Any alternative arguments are very welcome.