Show that $A:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$ where $A(e_n)=\lambda_ne_n$ is bounded. Let $C\subset\mathbb C$ be closed. As $\mathbb C$ is separable then so too is the subset $C$. This means that there exists a countable subset $\{\lambda_n:n\in\mathbb N\}\subset C$ dense in $C$.
In the first comment to the question here the claim is made that the operator $A:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$, the action of which is $A(e_n)=\lambda_ne_n$, is bounded. Here, $\{e_n:n\in\mathbb N\}\subset\ell^2(\mathbb N)$ is an orthonormal basis. This is what I want to show and it should be incredibly easy to demonstrate - I am overlooking something terribly obvious here, to my embarrassment.
Here is my working so far. Recall that in any Hilbert space, $\mathcal H$, with an orthonormal basis, any $x\in\mathcal H$ can be uniquely expressed in the form, $$x=\sum_{n=1}^\infty\langle x,e_n\rangle e_n.$$ Now, consider the quantity,
$$\begin{align*}
\|Ax\|&=\left\|A\left(\sum_{n=1}^\infty\langle x,e_n\rangle e_n\right)\right\|\\
&=\left\|\sum_{n=1}^\infty\langle x,e_n\rangle Ae_n\right\| &&\text{(assuming A linear)}\\
&=\left\|\sum_{n=1}^\infty\langle x,e_n\rangle \lambda_ne_n\right\|\\
&\le\sum_{n=1}^\infty\|\langle x,e_n\rangle \lambda_ne_n\| &&\text{(triangle inequality)}\\
&=\sum_{n=1}^\infty|\langle x,e_n\rangle| |\lambda_n|\|e_n\|\\
&\le\|x\|\sum_{n=1}^\infty |\lambda_n|\|e_n\|^2 &&\text{(Cauchy-Schwarz).}\\
\end{align*}$$
What I thought here to do was to bound $|\lambda_n|$ by $M:=\max_{n\in\mathbb N}\{\lambda_n\}$ so as to obtain,
$$\|x\|\sum_{n=1}^\infty |\lambda_n|\|e_n\|^2\le M\|x\|\sum_{n=1}^\infty \|e_n\|^2.$$
But as the norm of each $e_n\in\ell^2(\mathbb N)$ is unity the series we are left with diverges.
As to trying to show that the operator is bounded, what am I overlooking or failing to do correctly?
 A: It's not enough to have $C$ closed; you need it bounded, too. That is, you want $C$ compact. If the sequence $\{\lambda_n\}$ is not bounded, your $A$ will not be bounded. 
When dealing with series, the triangle inequality is very unlikely to give you sharp inequalities; you've gone too far. What you want here is Parseval's identity. Write $c=\sup_n|\lambda_n|$. Note also that you can only evaluate on finite sums, since a priori you don't know that $A$ is bounded. You have 
$$
\left\|\sum_{n=1}^M\lambda\langle x,e_n\rangle\,e_n\right\|^2=\sum_{n=1}^M|\lambda_n|^2\,|\langle x,e_n\rangle|^2\leq c^2\,\sum_{n=1}^M \,|\langle x,e_n\rangle|^2\leq c^2\,\|x\|^2. 
$$
As you can do this for all $M$ and all $x$, you get that $A$ is bounded on a dense subset of $H$, and thus extends to a bounded operator on all of $H$. 
A: You want to assume $C$ is compact, not just closed.  Then if $C$ is bounded by $M$, for any $x = (x_1, x_2, \ldots) \in \ell^2$, 
$A x = (\lambda_1 x_1, \lambda_2 x_2, \ldots)$ with
$$ \|A x \|^2 = \sum_{n=1}^\infty |\lambda_n|^2 |x_n|^2 \le M^2 \sum_{n=1}^\infty |x_n|^2 = M^2 \|x\|^2 $$
