Here are two examples from algebra where finiteness assumptions can be removed.

1. A finitely generated projective module over a local ring is free, but Kaplansky showed the result is true without the finite generatedness hypothesis: https://en.wikipedia.org/wiki/Kaplansky%27s_theorem_on_projective_modules.

2. The normal basis theorem says that for a finite Galois extension $L/K$, there is some $\alpha \in L$ such that the set of its $K$-conjugates $\{\sigma(\alpha) : \sigma \in {\rm Gal}(L/K)\}$ is a $K$-basis of $L$ (called a "normal basis"). If $L/K$ is an infinite Galois extension, then the normal basis theorem as described above does not make sense since $\{\sigma(\alpha) : \sigma \in {\rm Gal}(L/K)\}$ is a finite set for each $\alpha$ and thus could not be a $K$-basis when $L/K$ is infinite. Lenstra found a way to reformulate the definition of a normal basis so it does make sense for infinite Galois extensions and the normal basis theorem is then true in that setting. See https://pub.math.leidenuniv.nl/~lenstrahw/PUBLICATIONS/1985c/art.pdf. The basic idea is that the normal basis theorem for finite Galois extensions is about a comparison between ${\rm Map}(G,K)$ and $L$, where $G = {\rm Gal}(L/K)$ and ${\rm Map}(G,K)$ is the set of all maps $G \to K$. When $L/K$ is an infinite Galois extension, we should replace ${\rm Map}(G,K)$ with the set $C(G,K)$ of all continuous maps $G \to K$, where $G$ has the Krull topology and $K$ has the discrete topology. Lenstra's version of the normal basis theoem reduces to the usual version of the normal basis theorem when $L/K$ is finite since $C(G,K) = {\rm Map}(G,K)$ when $L/K$ is finite, as $G$ in that case has the discrete topology.

Here are some examples from algebra where finiteness assumptions can be removed. In the first two, the statement of the more general result is unchanged, but the third result has to be expressed in a new way in order to have a chance of being true without the finiteness condition.

1. From the classification of finitely generated modules over a PID, every submodule of a finitely generated free module over a PID is free. That consequence is also true without finite generatedness: every submodule of a free module over a PID is free. See https://math.stackexchange.com/questions/162945.

2. A finitely generated projective module over a local ring is free, but this result is also true without finite generatedness. See https://en.wikipedia.org/wiki/Kaplansky%27s_theorem_on_projective_modules.

3. The normal basis theorem says that for a finite Galois extension $L/K$, there is some $\alpha \in L$ such that the set of its $K$-conjugates $\{\sigma(\alpha) : \sigma \in {\rm Gal}(L/K)\}$ is a $K$-basis of $L$ (called a "normal basis"). If $L/K$ is an infinite Galois extension, then the normal basis theorem as described above does not make sense since $\{\sigma(\alpha) : \sigma \in {\rm Gal}(L/K)\}$ is a finite set for each $\alpha$ and thus could not be a $K$-basis when $L/K$ is infinite. Lenstra found a way to reformulate the definition of a normal basis so it does make sense for infinite Galois extensions and the normal basis theorem is then true in that setting. See https://pub.math.leidenuniv.nl/~lenstrahw/PUBLICATIONS/1985c/art.pdf. The basic idea is that the normal basis theorem for finite Galois extensions is about a comparison between ${\rm Map}(G,K)$ and $L$, where $G = {\rm Gal}(L/K)$ and ${\rm Map}(G,K)$ is the set of all maps $G \to K$. When $L/K$ is an infinite Galois extension, we should replace ${\rm Map}(G,K)$ with the set $C(G,K)$ of all continuous maps $G \to K$, where $G$ has the Krull topology and $K$ has the discrete topology. Lenstra's version of the normal basis theoem reduces to the usual version of the normal basis theorem when $L/K$ is finite since $C(G,K) = {\rm Map}(G,K)$ when $L/K$ is finite, as $G$ in that case has the discrete topology.