Let $(\lambda_n)_{n\in N}$ be an arbitrary sequence in $\mathbb{C}$ and hilbert space $\mathcal{H}_1 := l^2(\mathbb{N})$. Consider the operator $(S,D(S))$: \begin{align*} D(S) := \{x\in\mathcal{H}_1 : \exists N \in \mathbb{N}, x_n = 0 \text{ if } n \geq N\}, \hspace{20pt} Sx := (\lambda_nx_n)_{n\in\mathbb{N}}\,. \end{align*} Describe the closure of $S$.
I know that $S$ is continous, linear and therefore closed. Does $S = \bar{S}$ then apply immediately? Or do I have to do something else? Do I perhaps have to calculate $\bar{S}$ via the adjoint, so $\bar{S} = S^{**}$?
You don't know that $S$ is continuous. In fact, it will not not be continuous. For an easy example, take $\lambda_n=n$ for all $n$.
To look at the closure, we look at the graph of $S$: that is $$ \{(x,Sx):\ x\in D(S)\}. $$ Suppose now that $(x^m,Sx^m)\to (x,y)$ with $x^m\in D(S)$ for all $m$. Note that $$ y_n=\lim_m(Sx^m)_n=\lim_m\lambda_nx^m_n=\lambda_n\lim_m x^m_n=\lambda_n x_n. $$ Since $y\in\ell^2(\mathbb N)$, we have $$\tag1 \sum_n|\lambda_n|^2|x_n|^2<\infty. $$ Condition $(1)$ is also sufficient, so $$ D(\bar S)=\{x:\ \sum_n|\lambda_n|^2|x_n|^2<\infty\} $$ and $$ \bar S x=(\lambda_nx_n). $$