Convergent sequence of real numbers inside spectrum of operator I'm trying to solve the following.

Let $T$ be a compact and self-adjoint operator on a prehilbertian space.
Determine wether the spectrum of T contains a sequence of real numbers that converge to zero.

Some context:
By a prehilbertian space, we're referring to a vector space over a field $\mathbb{K}$, generally $\mathbb{R}$ or $\mathbb{C}$, that has an inner product. Every inner product naturally induces a norm so every prehilbertian space is also a normed vector space.
The resolvent of an operator $T$ is defined as $\rho(T) = \{\lambda\in\mathbb{C}:(\lambda Id-T) \ \text{has an inverse}\}$ and therefore the spectrum is $\sigma(T) = \mathbb{C}-\rho(T)$.
Finally, an operator $T:X\longrightarrow Y$ between normed spaces is compact if $T(B_X)$ is a relatively compact set in $Y$ (i.e. its closure is compact in $Y$), where $B_X$ is the unit ball in $X$.
My attempt at the proof
I'm having issues understanding what the restriction of not being a Hilbert space implies. Is the completeness of the space necessary to prove the claim? I mean, if we were on a Hilbert space, then by the spectral theorem there exists a Hilbert base and sequence of real (since $T$ is self-adjoint) numbers such that $Te_k = \lambda_ke_k$, with $\lambda_k\to 0$ as $k\to\infty$, therefore those real numbers are indeed eigenvalues, which are obviusly a subset of the spectrum. But what happens if it's not a Hilbert space? As far as I've been told, the result is true, but I don't know what results to use to come up with the answer.
 A: If $T$ is finite dimensional the conclusion is obvious as $\ker T$ is infinite dimensional. Hence  there is an infinite orthonormal system $e_n$ such that $Te_n=\lambda_ne_n,$ where $\lambda_n=0.$
Assume $T:\mathcal{H}_0\to \mathcal{H}_0$ is an infinite dimensional compact operator. Let $\mathcal{H}$ denote the completion of $\mathcal{H}_0.$ Then, by continuity, the operator $T$ extends uniquely to a bounded operator $\tilde{T}:\mathcal{H}\to \mathcal{H}.$ Moreover the operator $T^*$ extends uniquely to $(\tilde{T})^*,$ which follows from the formula
$$\langle T^*x,y\rangle = \langle x, Ty\rangle,\quad x,y\in \mathcal{H}_0$$
As $T^*=T$ then  $(\tilde{T})^*=\tilde{T},$ i.e. $\tilde{T}$ is self-adjoint on $\mathcal{H}.$
The operator $\tilde{T}$ is compact. Indeed assume $\{x_n\}$ is a bounded sequence in $\mathcal{H}.$ For every $n$ there exists $y_n\in \mathcal{H}_0$ such that $\|y_n-x_n\|<{1\over n}.$ Hence $\{y_n\}$ is a bounded sequence in $\mathcal{H}_0.$ By assumption $\{y_n\}$ contains a subsequence $\{y_{n_k}\}$ convergent to an element $y\in\mathcal{H}_0.$ Thus
$$x_{n_k}=y_{n_k}+(x_{n_k}-y_{n_k})\to y$$
If $\lambda\in \mathbb{C}\setminus \sigma(T)$ then the operator $\lambda I_{\mathcal{H}_0}-T$ admits an inverse, say $A_\lambda.$  Thus
$$A_\lambda(\lambda I_{\mathcal{H}_0}-T)=(\lambda I_{\mathcal{H}_0}-T)A_\lambda=I_{\mathcal{H}_0}$$
All operators above extend uniquely to bounded operators on $\mathcal{H},$ denoted by superscript $\tilde{}$, and
$$\tilde{A}_\lambda (\lambda I_{\mathcal{H}}-\tilde{T})=(\lambda I_{\mathcal{H}}-\tilde{T})\tilde{A}_\lambda=I_{\mathcal{H}}$$
Therefore $\lambda\in \mathbb{C}\setminus\sigma(\tilde{T}).$ In this way we have obtained $\sigma(\tilde{T})\subset \sigma({T}).$ As $\tilde{T}$ is infinite dimensional the set $\sigma(\tilde{T})$ is infinite and
$$\sigma(\tilde{T})=\{0\}\cup\{\lambda_n\}_{n=1}^\infty$$ where $\lambda_n\to 0.$
It is possible to give an example of an incomplete inner product space $\mathcal{H}_0$ and an infinite dimensional compact operator $T:\mathcal{H}_0\to \mathcal{H}_0.$
Let  $$\mathcal{H}_0=\left \{x\in \ell^2\,:\, \sum_{n=1}^\infty 4^n|x_n|^2<\infty\right\}\subset \ell^2=\mathcal{H}$$ and $$\tilde{T}x=\sum_{n=1}^\infty 2^{-n}x_ne_n$$ where $\{e_n\}_{n=1}^\infty$ denotes the standard orthonormal basis in $\ell^2.$ Observe that $\tilde{T}(\ell^2)\subset \mathcal{H}_0.$ Hence  the operator $T=\tilde{T}\mid_{\mathcal{H}_0}$ maps $\mathcal{H}_0$ into $\mathcal{H}_0.$
The operator $\tilde{T}$ is compact. We will show that $T$ is compact as well. To this end,
let $x^{(k)}$ be a bounded sequence in $\mathcal{H}_0\subset \ell^2.$ By the Banach-Alaoglu theorem there exists a subsequence $x^{(k_m)}$ convergent weakly to some $x\in \ell^2.$ By the compactness of $\tilde{T}$ we get $$Tx^{(k_m)}=\tilde{T}x^{(k_m)}\underset{m\to \infty}{\longrightarrow} \tilde{T}x\in \mathcal{H}_0$$
