Possible categorical reformulation for the usual definition of compactness

This is very far from being true. In fact, a compact space has your property iff it is finite.

To prove this, suppose $X$ is infinite and has your property. For each ultrafilter $U$ on $X$ converging to a point $x\in X$, let $A_{U,x}$ be a space with the same underlying set as $X$ such that a set is open iff it either does not contain $x$ or it contains $x$ and is in $U$. Then the identity maps $A_{U,x}\to X$ generate the topology on $X$ (this is just a restatement of the fact that a map is continuous iff it preserves convergence of ultrafilters). By assumption, there is a finite subfamily that generates the topology. Since $X$ is infinite, we can find an infinite subset $B\subset X$ which is not in any of the finitely many ultrafilters of the subfamily and does not contain any of the finitely many limit points of the subfamily. But then the complement of $B$ is open, as is every subset of $B$. This implies that $X$ is not compact, which is a contradiction.

(More generally, without assuming $X$ is compact, this implies that $X$ must have the topology generated by the convergence of finitely many ultrafilters. This implies (though this takes a little work to prove) that besides principal ultrafilters converging to their corresponding points, there are only finitely many pairs $(U,x)$ where $U$ is an ultrafilter on $X$ which converges to $x$. Conversely, if the topology on $X$ has only finitely many such nontrivial limits of ultrafilters, then $X$ does have your property, since there are only finitely many topologies strictly finer than the topology on $X$ and each one must be ruled out by a single $f_i$. Note in particular that contrary to what you say, $X$ does not need to be compact. For instance, if $X$ is infinite and discrete, then $X$ has the final topology for the empty family of maps, so it satisfies your property. The error in your argument is that $X$ can have the final topology without the images of the maps $f_i$ covering $X$.)

Note that for many spaces, you can see that your property fails much more easily. For instance, any metrizable space (or more generally, any countably generated space) has the final topology with respect to all inclusions of countable subspaces (because a set is closed iff it is sequentially closed). So any uncountable compact metric space is an easy counterexample.

I think the answer is no. For an example, let $X$ be the unit interval $[0,1]$ (with the usual topology).

For each $i \in \mathbb{N}^+$, let $Y_i = (1/i,1]$ with the usual topology. The obvious inclusion $f_i \colon Y_i \to X$ is continuous.

Let $Y_0$ be $[0,1]$ with an unusual topology: a set is open in $Y_0$ if and only if either it does not contain $0$ or it contains a neighborhood of $0$ in the usual topology. Any subset of $[0,1]$ that is open in the usual topology is open in $Y_0$, so the identity map $f_0 \colon Y_0 \to X$ is continuous.

It is obvious that no finite subfamily is final, because the union of a finite subset of $Y_1, Y_2, \ldots$ must miss some neighborhood $[0,\epsilon)$ of $0$. So every subset of $[0,\epsilon)$ that does not contain $0$ is open in the final topology for this subfamily.

On the other hand, the topology on $X$ is final for the entire family: suppose $f_i^{-1}(U)$ is open for every $i$.

• If $0 \notin U$, then $U = \bigcup_{i=1}^\infty (U \cap Y_i)$ is the union of open sets, and is therefore open.

• If $0 \in U$, then, since $U$ is open in the topology of $Y_0$, it must contain a neighborhood $[0,1/n)$ of $0$ in the usual topology. Then $U = [0,1/n) \cup ( U \cap Y_{2n})$ is the union of two open sets, and is therefore open.