My attempted Solution: We need to show that for a compact operator $T$ in a Banach space, the kernel of Id $T$, denoted as $\operatorname{ker}(\mathrm{Id}-T)$, is finite-dimensional.
The kernel of $\mathrm{Id}-T$ consists of all vectors $x$ in the Banach space for which $(\mathrm{Id}-$ $T) x=0$. Equivalently, these are vectors such that $T x=x$, meaning they are eigenvectors of $T$ corresponding to the eigenvalue 1 .
Since $T$ is a compact operator, it has the property of mapping bounded sets to relatively compact sets. In a Banach space, a relatively compact set has compact closure, which means every sequence in the set has a convergent subsequence.
Assume, for the sake of contradiction, that $\operatorname{ker}(\operatorname{Id}-T)$ is infinite-dimensional. This would imply that there exists an infinite sequence of linearly independent vectors $\left\{x_n\right\}$ in $\operatorname{ker}(\operatorname{Id}-T)$. We can normalize these vectors so that their norm is 1 , making $\left\{x_n\right\}$ a bounded sequence.
Given that $T$ is compact, the image under $T$ of any bounded sequence has a convergent subsequence. Therefore, the sequence $\left\{T x_n\right\}$ must have a convergent subsequence. Let's denote this convergent subsequence by $\left\{T x_{n_k}\right\}$.
However, since each $x_n$ is in the kernel of $\operatorname{Id}-T$, we have $T x_n=x_n$ for all $n$. Hence, the sequence $\left\{x_{n_k}\right\}$, being a subsequence of $\left\{x_n\right\}$, is also convergent.
This leads to a contradiction because a sequence of linearly independent vectors in a normed space cannot converge. This contratiction suggests that our initial assumption that $\operatorname{ker}(\mathrm{Id}-T)$ is infinite-dimensional must be false.
-- I edited it according to the comment, please read it now!
It is not true that a sequence of linearly independent unit vectors cannot be convergent. For example, in $\ell^1$ with the standard basis $v_j$ take the sequence $x_j = (j v_1 + v_j)/(j+1)$. These are linearly independent and converge to $v_1$.