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. $\square$
Here's a proof I learned from some notes of Richard Melrose. I just noticed another answer was posted while I was typing. This uses a different characterization of compactness so hopefully it is interesting for that reason.
First, I claim that a set $K\subset H$ of a Hilbert space is compact if and only if it is closed, bounded, and satisfies the equi-small tail condition with respect to any orthonormal basis. This means that given a basis $\{e_k\}_{k=1}^\infty$, for any $\varepsilon>0$ we can choose $N$ large enough such that for any element $u\in H$,
$$\sum_{k>N} |\langle u , e_k \rangle |^2 < \varepsilon.$$
The main point here is that this condition ensures sequential compactness. The proof is a standard "diagonalization" argument where you choose a bunch of subsequences and take the diagonal. This is done in detail on on page 77 of the notes I linked.
With this lemma in hand, the proof is straightforward. I repeat it from page 80 of those notes. Fix a compact operator $T$. By definition [this is where we use a certain characterization of compactness] the image of the unit ball $T(B(0,1))$ is compact. Then we have the tail condition that, for given $\varepsilon$, there exists $N$ such that
$$\sum_{k>N} |\langle Tu , e_k \rangle |^2 < \varepsilon.$$
for any $u$ such that $\| u \| < 1$.
We consider the finite rank operators
$$T_nu = \sum_{k\le n} \langle Tu , e_k \rangle e_k.$$
Now note the tail condition is exactly what we need to show $T_n \rightarrow T$ in norm. So we're done.
Best Answer
True if $\lambda \neq 0$. We may suppose $\|x_n\|=1$ for all $n$. We have $\|\lambda x_n-Tx_n\| \leq \|T-T_n\| \to 0$. Since $T$ is compact there is subsequence of $(Tx_n)$ which converges in the norm. But then the corresponding subsequence of $(x_n)$ itself converges. If $x$ is the limit of this subsequence then $Tx=\lambda x$. Also, $\|x\|=1$, so $x \neq 0$.