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.
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.