Consider any diagram of finite $T_0$-spaces where the spaces involved are at most countably many. So we have $X_0, X_1, \ldots, X_n, \ldots$ each of which is $T_0$ and finite and some continuous maps between them. Suppose $L$ is the limit of that diagram in the category $\text{Top}$.
Is $L$ locally compact (or even compact)?
Here we take the definition of locally compact as definition 2.1 on nLab: every point has a neighbourhood base consisting of compact subspaces. In other words: a topological space $X$ is locally compact if for every $x \in X$ and every open neighbourhood $U$ of $x$ there is some compact neighbourhood $K$ of $x$ with $K \subseteq U$.
Yes, $L$ is always compact. More generally, the limit of any diagram of finite spaces is compact. To prove this, first note that if all the spaces are discrete, then the limit is a closed subset of the product of all the spaces, which is compact by Tychonoff's theorem. If our spaces aren't discrete, observe that the topology on their limit is coarser than the topology on the limit of the diagram of discrete spaces with the same underlying sets, and thus is also compact.
Note that for non-Hausdorff spaces, compactness does not imply local compactness (see Find an example of a compact space which is not locally compact.). However, a limit of finite spaces is always also locally compact, since the basic open neighborhoods are all compact (they are also limits of diagrams of finite spaces, obtained by just restricting your original diagram to appropriate subspaces of each space).
Topological spaces which are limits of diagrams of finite $T_0$ spaces are known as spectral spaces and have many other special properties. In particular, they are exactly those topological spaces which can be the spectrum of a commutative ring.