Clearly the natural map $|\mathbb A^{n+m}| \to |\mathbb A^n| \times |\mathbb A^m|$ is not a homeomorphism unless $n = 0$ or $m = 0$ (it's not even a bijection, because of points like $(x-y) \in \operatorname{Spec} k[x] \otimes k[y]$). We show that in fact there is no homeomorphism, using the following ad hoc notion of dimension:
Definition. Let $X$ be a topological space. Then the intersection dimension $\operatorname{int. dim}(X)$ of $X$ is the supremum of all $n$ for which there exist irreducible closed sets $V_0,\ldots,V_n \subseteq X$ such that
- any intersection $V_I = \bigcap_{i\in I} V_i$ for $I \subseteq \{0,\ldots,n\}$ nonempty is either irreducible or empty;
- the intersection $V_I$ for any $I \subseteq \{0,\ldots,n\}$ with $|I| = n$ is exactly one point;
- the intersection $V_{\{0,\ldots,n\}}$ is empty.
I feel that I have seen some variant of this somewhere before (probably under a different name), but I do not recall where. References are welcome!
Remark. If there exist sets $V_0,\ldots,V_n$ with properties 1, 2, and 3 above, then the same is true for any $m \leq n$. Indeed, we may take the sets $V_0, \ldots, V_{m-1}, V_{\{m,\ldots,n\}}$.
Lemma 1. Let $k$ be a field. Then $\operatorname{int. dim}(\mathbb A^n_k) = n$.
Proof. The hyperplanes $V(x_1+\ldots+x_n), V(x_1), \ldots, V(x_n)$ show that $\operatorname{int. dim}(\mathbb A^n_k) \geq n$. Conversely, let $V_0, \ldots, V_m$ be irreducible closed subsets satisfying 1, 2, and 3 above. Conditions 2 and 3 imply that no $V_i$ is contained in any intersection of the others, so by induction on $|I|$ the dimension of $V_I$ is at most $n-|I|$ (because all intersections are irreducible). Taking $|I| = m$ gives $m \leq n$, showing that $\operatorname{int. dim}(\mathbb A^n_k) \leq n$. $\square$
More generally, this argument shows that if $X$ is a topological space of dimension $n$, then $\operatorname{int. dim}(X) \leq \dim X$. It's not clear even for smooth proper varieties that the two are equal (in the projective case we might be able to use some Bertini argument, using tangency to reduce the intersections to one point).
Lemma 2. Let $X$ and $Y$ be topological spaces with a closed point. Then
$$\operatorname{int. dim}(X \times Y) = \max(\operatorname{int. dim}(X), \operatorname{int. dim}(Y)).$$
Proof. By the Lemma of this post, a subset $Z \subseteq X \times Y$ is closed and irreducible if and only if $Z = V \times W$, with $V \subseteq X$ and $W \subseteq Y$ closed and irreducible.
Let $m$ be the maximum of the intersection dimensions of $X$ and $Y$; without loss of generality let's say $m = \operatorname{int. dim}(X)$. Then there exist sets $V_0,\ldots,V_m \subseteq X$ satisfying properties 1, 2, and 3 of the definition above. Choose a closed point $y \in Y$, and set $W_i = \{y\}$ for all $i$. Then the sets $Z_i = V_i \times W_i$ are closed and irreducible. Moreover, the finite intersections $Z_I = \bigcap_{i \in I} Z_i$ equal $V_I \times W_I$. Since properties 1, 2, and 3 hold for the $V_i$ and properties 1 and 2 for the $W_i$, this shows properties 1, 2, and 3 for the $Z_i$. Therefore, $\operatorname{int. dim}(X \times Y) \geq m$.
Conversely, if $Z_0, \ldots, Z_n$ satisfy properties 1, 2, and 3, then we must have $Z_i = V_i \times W_i$ for some irreducible closed subsets $V_i \subseteq X$, $W_i \subseteq Y$. Again we have $Z_I = V_I \times W_I$, so both $V_i$ and $W_i$ satisfy 1 and 2. Since $Z_{\{0,\ldots,n\}} = \varnothing$, we conclude that either $V_{\{0,\ldots,n\}} = \varnothing$ or $W_{\{0,\ldots,n\}} = \varnothing$. Hence, either the $V_i$ or the $W_i$ satisfy the conditions above, so $\operatorname{int. dim}(X \times Y) \leq m$. $\square$
Corollary. There exists a homeomorphism $|\mathbb A^{n+m}_k| \stackrel\sim\to |\mathbb A^n_k| \times |\mathbb A^m_k|$ if and only if $n = 0$ or $m = 0$.
Proof. By Lemma 1 and Lemma 2, we have
$$\operatorname{int. dim}(|\mathbb A^n_k| \times |\mathbb A^m_k|) = \max(n,m),$$
whereas
$$\operatorname{int. dim}(|\mathbb A^{n+m}_k|) = n + m.$$
These numbers are equal if and only if $n = 0$ or $m = 0$. Clearly the stated homeomorphisms exist in this case. $\square$
We can actually mix and match fields, i.e. consider things like $|\mathbb A^{100}_{\mathbb C}| \times |\mathbb A^{17}_{\bar{\mathbb F_3}}|$. The same proof shows that this cannot be homeomorphic to any space of the form $|\mathbb A^{117}_k|$ for a field $k$.
Best Answer
The comments of @freakish have given a false impression. The product topology makes all projections $\pi_j:\prod_iX_i\to X_j$ continuous, and is the coarsest topology with this property.
The result is that there is a sub-basis of the open sets, consisting of the “cylindrical” sets $\pi_j^{-1}U_j$ where $U_j$ is an open set of one of the factors. So, for instance, $\Bbb C\times\{w_0\}$ is closed in the product topology, as is each $\{z_0\}\times\Bbb C$. Notice that limiting the closed sets to the cofinite subsets of $\Bbb C\times\Bbb C$ does not make either projection continuous. To get the product topology from the subbase, you allow finite intersections of the cylindrical sets, and arbitrary unions of these.
For a Zariski-closed subset, take any obliqilue line $w=mz+b$ for $m\ne0$, or, much more realistically, a circle $z^2+w^2=1$.