Combing Compact Operators and Pre-Compactness

Question

Background

I have had a lecture in my class on what it means for an operator to be compact i.e. for any bounded B, $\overline{T(B)}$ is compact. Now I have also had a theorem about classifying compactness in metric spaces:

1. In a metric space a sequence is compact iff it is complete and totally bounded.

And finally I had a lecture on when are sets totally bounded/precompact in particular spaces, like:

1. A subset $\mathcal{G} \subset l_{p}$, where $1 \leq p<\infty$, is totally bounded iff (i) it is pointwise bounded, i.e. $\forall j \in \mathbb{N}$ $$ \begin{array}{l} \sup _{x \in \mathcal{G}}\left|x_{j}\right|<\infty \\ \end{array} $$ and (ii) for every $\varepsilon>0$ there is some $n$ so that, for every $x \in \mathcal{G}$ $$ \sum_{k>n}\left|x_{k}\right|^{p}<\varepsilon^{p} $$

2. Let $\mathcal{H}$ be a separable Hilbert space. Suppose $\mathcal{D} \subset \mathcal{H}$ is a bounded set. If for some orthonormal basis $\left\{e_{n}\right\}_{n \in \mathbb{N}}$ of $\mathcal{H}$ we have $$ \forall \varepsilon>0 \quad \exists N \in \mathbb{N} \quad \sum_{n=N}^{\infty}\left|\left\langle f, e_{n}\right\rangle\right|^{2}<\varepsilon^{2} $$ for all $f \in \mathcal{D}$, then $\mathcal{D}$ is precompact in $\mathcal{H}$.

3. (Arzela-Ascoli). Let $\Omega$ be a compact topological space. Then a subset $\mathcal{F} \subset \mathcal{C}(\Omega)$ is totally bounded in the supremum norm $i f f$ (i) it is pointwise bounded, i.e. $\forall x \in \Omega$ $$ \sup _{f \in \mathcal{F}}|f(x)|<\infty $$ and (ii) it is equicontinuous, i.e. $\forall x \in \Omega$ and $\forall \varepsilon>0$ there exists a neighborhood $V$ of $x$ s.t. $\forall y \in V|f(x)-f(y)|<\varepsilon$ for all $f \in \mathcal{F}$ Compactness in $L_{p}\left(\mathbb{R}^{n}\right), 1 \leq p<\infty$, spaces.

Question

I'm not sure how to tie these topics together i.e. if I have an operator what are usually the steps I have to take to show it is compact. Would it be enough for me to take a bounded set and show T(B) is precompact or would it be enough to show T(B) is totally bounded? Would I need a condition that the codomain is Banach (for the closure to be compact)? What are usually the steps we take to show that an operator is compact? I'm sorry if this is a bit vague I'm just a bit lost. How to relate this topic of pre-compact and total boundedness?