You didn't say that but I'm assuming you also want $T$ to be densely defined.
Let us write $R_\lambda=(T-\lambda)^{-1}$ for $\lambda\in\rho(T)$ to denote the resolvent. If $R_\lambda$ is compact for some $\lambda$, then $T$ is said to have compact resolvent (just to give you a term to search for more information).
The statements that you make are very well-known facts about operators with compact resolvent.
For your convenience I will prove them below.
Assume that $R_{\lambda_0}$ is compact for some $\lambda_0\in\rho(T)$.
Claim 1. $R_\lambda=(T-\lambda)^{-1}$ is compact for every $\lambda\in\rho(T)$.
Proof. Recall the resolvent relation:
$$R_\lambda-R_{\lambda_0} = (\lambda-\lambda_0)R_\lambda R_{\lambda_0}$$
Adding $R_{\lambda_0}$ on both sides we get
$$R_\lambda = ( (\lambda-\lambda_0)R_\lambda + I) R_{\lambda_0}.$$
Thus, we have written $R_\lambda$ as composition of a bounded operator and a compact operator and therefore it is compact. $\square$
Claim 2. $\sigma(T)$ is a discrete set (countable and no accumulation points) and consists only of eigenvalues.
Proof. Set $S=R_{\lambda_0}$.
The assumptions is that $S$ is compact. We claim that this implies that $\sigma(S^{-1})$ is a discrete set of eigenvalues (which immediately implies the same for $\sigma(T)$ since $S^{-1}=T-\lambda$).
Because $S^{-1}$ is invertible, we have $0\not\in\sigma(S^{-1})$. Furthermore, by the spectral theorem, $\sigma(S)$ consists of countably many non-zero eigenvalues with the only possible accumulation point being $0$.
Observe that $\lambda\not=0$ is an eigenvalue of $S$ if and only if $\lambda^{-1}$ is an eigenvalue of $S^{-1}$ (check that!).
Next, observe that $\sigma(S^{-1})$ consists only of eigenvalues. To see this, say $0\not=\lambda$ is not an eigenvalue of $S$. Then $\lambda\in\rho(S)$. So $S-\lambda$ is bounded and invertible. This implies that $S^{-1}-\lambda^{-1}$ is also bounded and invertible:
$$S-\lambda = \lambda S(\lambda^{-1}-S^{-1})$$
$$(S^{-1}-\lambda^{-1})^{-1} = -\lambda (S-\lambda)^{-1}S$$
Thus $\lambda^{-1}\in \rho(S^{-1})$.
Putting these observations together we have proved that $$\sigma(S^{-1})=\sigma_p(S^{-1}) = \{\lambda^{-1}\,:\,\lambda\in\sigma(S)\setminus\{0\}\}$$
which is a discrete set. $\square$
The definition of eigenvalue is about operators. We use it on matrices because we can see them as operators. I say this because the definition of eigenvalue survives similarity, which means it is independent of basis, and thus it is independent of the form of the operator as a matrix.
Namely, if $Ax=\lambda x$, then for any invertible $S$ we have $(SAS^{-1})Sx=\lambda Sx$. So if $\lambda$ is an eigenvalue of $A$, it is also an eigenvalue of $SAS^{-1}$, so it is still an eigenvalue of $A$ in any basis.
In infinite dimension things change, first as you mention because the useful notion of spectrum goes beyond that of eigenvalue. Second, because there is no immediate way to define the notion of eigenvalue of an infinite matrix other than as an operator (i.e., $Ax=\lambda x$). There is no general infinite-dimensional analog of $\det (A-\lambda I)=0$.
Best Answer
If $0$ is not in the spectrum of $A$, then $A$ is invertible with inverse $B$. Since the product of compact operators is compact, the identity $I = AB$ is compact. But this is impossible if $A$ acts on an infinite-dimensional Banach/Hilbert space, because the closed unit ball is not compact.