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This is very far from being true. In fact, a 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.

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.

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.

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

To prove this, let $X$ be a space with your property, and 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). If there is a finite subcollection that generates the topology, then that means there are finitely many ultrafilters whose convergence determines the topology. But if $X$ is infinite, this means that $X$ cannot be compact, because any nonprincipal ultrafilter that is not on our finite list will not have a limit. But as you observed, $X$ must be compact, and therefore $X$ is finite.

Note that for most reasonably nice 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.

This is very far from being true. In fact, a 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.

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.

This is very far from being true. 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 a counterexample.

In fact, by a generalization of the construction in Dave's answer, we can see that spaces satisfying your condition are extremely restricted. Given any space $X$ and a point $x\in X$, let $A_x$ be $X$ topologized such that a set is open iff it either does not contain $x$ or it contains a neighborhood of $x$ in $X$. Then $X$ has the final topology with respect to the identity maps $A_x\to X$ over all points $x\in X$. But if it has the final topology with respect to some finite subcollection, then there must be a cofinite subset of $X$ that is discrete! With a bit more work, it is not hard to see that if $X$ is Hausdorff, it must be finite.

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

To prove this, let $X$ be a space with your property, and 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). If there is a finite subcollection that generates the topology, then that means there are finitely many ultrafilters whose convergence determines the topology. But if $X$ is infinite, this means that $X$ cannot be compact, because any nonprincipal ultrafilter that is not on our finite list will not have a limit. But as you observed, $X$ must be compact, and therefore $X$ is finite.

Note that for most reasonably nice 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.