Excuse my notation, it carries from physics.
Suppose we have an operator $A$. The eigenvalue $\lambda$ is said to be $g$-fold degenerate if there exists g linearly independent eigenvectors with eigenvalue $\lambda$. So we can define the set of these linearly independent kets $\{|\psi\rangle\}_{i=1,...,g}$.
Furthermore, we can form a $g$-dimensional eigensubspace $\mathcal{E}$ from this set. Now, consider a ket $|\chi\rangle$ that lives in $\mathcal{E}$; then, we can express it as a linear combination of our set of kets such that $$|\chi\rangle = \sum_{i=1}^g \alpha_i\ |\psi\rangle_i.$$
Now, $$A\ |\chi\rangle=\sum_{i=1}^g \alpha_i\ A\ |\psi\rangle_i=\lambda \sum_{i=1}^g \alpha_i\ |\psi\rangle_i=\lambda |\chi\rangle.$$
Thus, we see that $|\chi\rangle$ too has an eigenvalue of $\lambda$. In fact, we have infinite number of kets that can be written as a linear combination of the set $\{|\psi\rangle\}_{i=1,...,g}$.
So why isn't $\lambda$ infinitely-degenerate? From this argument, it seems that if $\lambda$ has degeneracy greater than 1, then it would be infinitely degenerate since we can write infinitely many kets that have the same eigenvalue.
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$\begingroup$There are indeed infinitely many eigenvectors, but not infinitely many linearly independent ones. The dimension $g$ of the subspace is the largest possible number of linearly independent vectors in the subspace.
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