I'm asking this question out of curiosity, as I was unable to come to a conclusion.
Consider the affine space over an algebraically closed field, for concreteness we can work with $\mathbb{C}^{n}$. Let's restrict ourselves to the closed points, ie. we're working with the spectrum of maximal ideals. What is the homotopy type of this space..?
I'd be happy with only knowing the weak homotopy type and I have some suspicion that $\mathbb{C}^{n}$ might be weakly contractible, but I was unable to prove it. Since the Zariski topology $\mathbb{C}^{n}$ is strictly weaker than the usual one, it is of course easy to construct homotopies between maps $S^{n} \rightarrow \mathbb{C}^{n}$, as all the homotopies in the usual topology will work. However, the problem lies in the fact that, in principle, there might be simply much more continous maps from the sphere.
One might also try to consider the "radial contraction" map $r: \mathbb{C}^{n} \times \mathbb{C} \rightarrow \mathbb{C}^{n}$ given by $r(v, t) = tv$ which is also continous in the Zariski topology and then restrict that to $\mathbb{C}^{n} \times [0, 1]$ to prove contractibility directly. However, Zariski topology on the product is stronger than the product topology, so it doesn't work.
Edit: I have considered the case of $\mathbb{C}$ and it seems it is indeed contractible, but maybe I'm missing something obvious. We're looking for a map $H: \mathbb{C}^{1} \times I \rightarrow \mathbb{C}^{1}$ that restricts to the identity on $\mathbb{C}^{1} \times \{ 1 \}$ and to the constant map on $\mathbb{C}^{1} \times \{ 0 \}$. Note that it's enough that the preimages of points are closed, since the closed sets in $\mathbb{C}^{1}$ are finite or the whole space. Choose a bijection $\phi: \mathbb{C}^{1} \times (0, 1) \simeq \mathbb{C}^{1}$ and put $H(z, t) = \phi(z, t)$ for $t \in (0, 1)$. Then the preimage of every in $\mathbb{C}^{1}$ under $H$ is some closed subset of $\mathbb{C}^{1} \times \{ 0, 1 \}$ plus a single point, so it is closed.