This is actually fairly simple. Let $I$ be a radical ideal in $k[x_1, \dots, x_n]$, $k$ algebraically closed, then it corresponds to a unique algebraic set $V \subseteq \mathbb{A}^n(k)$, $V=Z(I)$.
So then $V = V_1 \cup \dots \cup V_k$, where $V_k$ are algebraic varieties corresponding uniquely to prime ideals $I_1, \dots, I_k$. Thus $Z(I) = Z(I_1) \cup \dots \cup Z(I_k)$. Now one can show that when taking the union of algebraic sets corresponding to arbitrary ideals, that $$Z(I_1)\cup \dots \cup Z(I_k) = Z(I_1 \cdots I_k) $$ where the product is the product of ideals. Also one has in general that $Z(J)= Z(\sqrt{J})$, so that $$Z(I_1)\cup \dots \cup Z(I_k) = Z(I_1 \cdots I_k)=Z\left(\sqrt{I_1 \cdots I_k}\right) $$ Now in general one only has that $I_1\cdots I_k \subseteq I_1 \cap \dots \cap I_k$, but taking the radical of the product allows us to conclude (see Lemma 1.7 here and then use induction): $$Z\left(\sqrt{I_1 \cdots I_k}\right)= Z\left( \sqrt{I_1 \cap \dots \cap I_k} \right) = Z\left( \sqrt{I_1} \cap \dots \cap \sqrt{I_k} \right). $$ However, since the $V_i$ are algebraic varieties, all of the $I_i$ were prime, thus radical, i.e. for all $i$ we have $\sqrt{I_i}=I_i$, therefore we have actually shown that: $$Z(I_1) \cup \dots \cup Z(I_k) =Z\left(\sqrt{I_1 \cdots I_k}\right) = Z(I_1 \cap \dots \cap I_k) $$ Now since $\sqrt{I_1 \cdots I_k} = I_1 \cap \dots \cap I_k$, the former obviously being radical since taking the radical of an ideal is idempotent operation, the latter must also be a radical ideal, and therefore our one-to-one correspondence between zero sets and radical ideals (from the Strong Nullstellensatz) applies, in particular we have that $$Z(I)=Z(I_1)\cup \dots \cup Z(I_k)=Z(I_1 \cap \dots \cap I_k) \iff I = I_1 \cap \dots \cap I_k.$$ Thus starting with an arbitrary radical ideal in $k[x_1,\dots,x_n]$, we have returned a (in fact unique) intersection of prime ideals equaling it, thereby proving this special case of the Lasker-Noether theorem. $\square$
Proof of the lemma (in case the link goes dead in the future)
The fact that $\sqrt{I \cdot J} \subseteq \sqrt{I \cap J}$ follows from the fact that $I \cdot J \subseteq I \cap J$.
If $ a \in \sqrt{I \cap J}$ then $a^n \in I \cap J$ for some $n\in \mathbb{N}$, thus both $a^n \in I$ and $a^n \in J$ are true, and thus $a \in \sqrt{I} \cap \sqrt{J}$. Thus $\sqrt{I \cap J} \subseteq \sqrt{I} \cap \sqrt{J}$.
Now if $a \in \sqrt{I} \cap \sqrt{J}$, then $a^n \in I$, $a^m \in J$, for some $m,n \in \mathbb{N}$. Therefore $a^{m+n} \in I \cdot J$, so $\sqrt{I} \cap \sqrt{J} \subseteq \sqrt{I \cdot J}$. $\square$
The correspondence between ideals and closed subsets reverses inclusion. Therefore, the expected ideal should not be the ideal $I^+$ generated by homogeneous parts of generators of $I$, as such ideal would contain $I$, but rather
$$\mathrm{Cone}(I)=I^{-}:=(\{f \in I\; |\;f \text{ is homogeneous}\}) \subseteq I.$$
Indeed, as $I^-$ is the biggest homogeneous ideal contained in $I$, $V(I^-)$ is the smallest conical subset containing $V(I)$, which is $\mathrm{Cone}(V(I))$.
(If you prefer to talk strictly about varieties, i.e. $I$ is assumed to be a radical ideal, note that then $I^-$ is radical too, so it works out.)
Best Answer
You can look at $V(x^2+1)$ in $\mathbb{R}$, ie the empty set. $x^2+1$ is a radical ideal, since $\mathbb{R} [x]/ (x^2+1)$ is isomorphic to $\mathbb{C}$, a reduced ring. You can also look at $V(\mathbb{R}[x])$.