The easiest way to reduce to the uncountable case may be as follows. Let $I$ be an ideal of $k[X_1,...,X_d]$ which does not contain $1$. Let $P_1,\dots,P_r$ be a generating family of $I$.
Let $A=k^{\mathbf N}$ and let $m$ be a maximal ideal of $A$ which contains the ideal $N=k^{(\mathbf N)}$ of $A$. Then $K=A/m$ is an algebraically closed field which is has at least the power of the continuum. (Alternative description: let $K$ be an ultrapower of $k$, with respect to a non-principal ultrafilter.)
Lemma. For $i\in\{1,\dots,r\}$, let $a_i=(a_{i,n})\in A$. Assume that $(\bar a_1,\dots,\bar a_r)=0$ in $K^r$. Then the set of $n\in\mathbf N$ such that $(a_{1,n},\dots,a_{r,n})=0$ is infinite.
Proof. Assume otherwise. For every $n$ such that $(a_{1,n},...,a_{r,n}) \neq 0$, choose $(b_{1,n},\dots,b_{r,n})$ such that $\sum a_{i,n}b_{i,n}=1 $, and let $b_i=(b_{i,n})_n\in A$. Then $\sum a_i b_i - 1 $ belongs to $N^r$, hence $\sum \bar a_i \bar b_i=1$. Contradiction.
Thanks to the lemma, one proves easily that the ideal $I_K$ of $K[X_1,...,X_d]$ generated by $I$ does not contain $1$. By the uncountable case, there exists $x=(x_1,...,x_d)\in K^d$ such that $P_j(x_1,...,x_d)=0$ for every $j$. For every $i$, let $a=(a_{n})\in A^d$ be such that $\bar a=x$. By the lemma again the set of integers $n$ such that $P_j(a_n) \neq 0$ for some $j$ is finite. In particular, there exists a point $y\in k^d$ such that $P_j(y)=0$ for every $j$.
