I want to show that $\operatorname{Spec}\mathbb{Z}$ can be written as an inverse limit of finite $T_0$-spaces. First off, $\operatorname{Spec}\mathbb{Z} = \{(0), (2),(3),(5),...\}$, so the closed sets in the Zariski topology are precisely the points $(p)$ with $p \neq 0$ as well as finite unions of these points. Since prime numbers are countable, we could think of $\operatorname{Spec}\mathbb{Z}$ as the set $\mathbb{N} \cup \{\infty\}$, where $\{\infty\}$ corresponds to the generic point $(0)$, and the open sets $A \subseteq \mathbb{N} \cup \{\infty\}$ are precisely of the form $A= \tilde{A} \cup \{\infty\}$, where $\tilde{A} \subseteq \mathbb{N}$ is open in the cofinite topology, i.e. $\mathbb{N}\setminus \tilde{A}$ is a finite set. My idea now was consider finite spaces $X_i = \{1, ... , i\}$, with some topology that looks similar to the cofinite topology. We want $\{i\}$ to be the generic point of $X_i$, so maybe we can define maps $$f_{i,i+1} : X_{i+1} \rightarrow X_i, 1 \mapsto 1, a \mapsto a-1, a \neq 1$$ This way, the generic point $i+1$ of $X_{i+1}$ gets mapped to the generic point $i$ of $X_i$, and every point gets mapped to a non-generic point. Now, for $i \leq j$ we can easily define maps $f_{i,j}$ by $f_{i,i} = id$, and $f_{i,j}$ as the compositions of above maps. This defines an inverse system of topological spaces (assumed that these maps are continuous), hence we can consider $$X := \lim_{\leftarrow} X_i = \left\{(x_1, x_2, ...) \in \Pi_{i \in \mathbb{N}} X_i : x_i = f_{i,i+1}(x_{i+1})\right\}$$ We can see that the elements of $X$ are of the form $$X = \{(1,2,3,4,5,...), (1,1,2,3,4, ...), (1,1,1,2,3,4,...), ... , (1,1,1,...)\} \\ =\{(x_1, x_2,...) : \exists L \in \mathbb{N}\cup\{\infty\} \text{ s.t. } x_k = 1 \text{ }\forall k\leq L, x_k = k+1-L \text{ } \forall k>L\} \\ =: \{A_1, A_2, ... , A_{\infty}\} $$ This looks somewhat similar to $\operatorname{Spec}\mathbb{Z}$, I would love to just map $A_\infty \mapsto (0)$, $A_1 \mapsto (2)$, $A_2 \mapsto (3)$ etc., but we require that this map is continuous as well. The topology on $X$ is given by the product topology, i.e. $U \subseteq X$ is open if it can be written as union of $\Pi_i U_i$, where the $U_i \subseteq X_i$ are open, and $U_i = f_{i,i+1} (U_{i+1})$. Now however, I don't know what topology to pick on the $X_i$. We need the maps $f_{i,j}$ to be continuous, so I thought about picking the following topology on $X_i$: $A \subseteq X_i$ open iff $A=\tilde{A} \cup \{i\}$, where $\tilde{A} \subseteq \{1, ... , i-1\}$. This looks somewhat similar to the cofinite topology but we have a finite space here, so it doesn't really work. First off, the $f_{i,j}$ are easily seen to be continuous, but $U_i :=\{1\} \cup \{i\}$ is always open in $X_i$, which means that $U = \Pi_i U_i$ is open in $X$, but this corresponds to $\{A_1, A_\infty\}$ open, which we do not want since $\{(0), (2)\}$ is not open in $\operatorname{Spec}\mathbb{Z}$.
What else could I try here? It seems like that the topology on $X_i$ needs to have bigger and bigger open sets so that in the limit we get infinitely big non-empty open sets, and also otherwise some finite union of the $A_i$ would be open. But these open sets also have to be stable under intersection, so in some way, they need to be somehow included in each other, so intersections cannot become arbitrarily small. Also, we need to guarantee continuity of the maps $f_{i,j}$.
Your idea works if you just tweak it so that the two points that get collapsed together by $f_{i,i+1}$ get sent to the generic point instead of to $1$. That is, define it instead by $a\mapsto a$ for $a\leq i$ and $i+1\mapsto i$. The inverse limit will then have one point corresponding to each point of $\mathbb{N}$ and one point that corresponds to the generic points in each $X_i$, and it is easy to check that it has the topology you want.
To motivate this a bit, the spaces in the inverse system should be finite spaces that $\mathbb{N}\cup\{\infty\}$ maps to. So, instead of just trying to guess an inverse system of finite spaces that could work, you can look specifically at finite quotients of $\mathbb{N}\cup\{\infty\}$. The most obvious way to get a finite $T_0$ quotient of $\mathbb{N}\cup\{\infty\}$ is to collapse everything except $\{1,\dots,i-1\}$ to a single point. These quotients naturally form an inverse system by the maps used in the previous paragraph.