I'd be tempted to use a sequence of itemize and enumerate environment, using the enumitem package to help with formatting
In the code below I've used the resume feature provided by the enumitem package to help keep the numbering of the enumerate environment without having to hardcode any of the numbers.
I've also used the optional argument of the \item command, which can be used, for example
\item[$\Vdash$]
If you plan to use this a lot though, you could always make this into a \newcommand
\documentclass{article}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{enumitem}
% global settings for enumerate environment
\setlist[enumerate]{label*=(\arabic*),leftmargin=0mm}
% itemize environment, when at nested depths 2, 3, 4
\setlist[itemize,2]{label=\textbullet,leftmargin=1cm}
\setlist[itemize,3]{label=\textbullet,leftmargin=1cm}
\setlist[itemize,4]{label=\textbullet,leftmargin=1cm}
\begin{document}
\begin{itemize}
\item Show that $f$ is continuous at $x_0$, when
\begin{enumerate}
\item\label{item:range} $f(x)=\sqrt{x}$ when $x\geq 0$, and
\item\label{item:pos} $x_0>0$
\end{enumerate}
\item[$\Vdash$] $f$ is continuous at $x_0$
\item[$\equiv$] $\{\text{definition of continuity}\}$
$(\forall\epsilon>0\cdot \exists\delta>0\cdot\forall x\geq 0\cdot |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\epsilon)$
\item[$\equiv$] $\{\text{choose arbitrary }\epsilon \text{generilization rule}\}$
\begin{itemize}
\item Show that $(\exists\delta>0\cdot\forall x\geq 0\cdot |x-x_0|<\delta\Rightarrow |f(x)-f(x_0)|<\epsilon)$, when
\begin{enumerate}[resume]
\item $\epsilon>0$
\end{enumerate}
\item[$\Vdash$] $\{\text{find a suitable value for } \delta\text{ witness rule}\}$
\begin{itemize}
\item Show that $\delta>0\wedge(\forall x\geq 0\cdot|x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\epsilon)$, when
\begin{enumerate}[resume]
\item $\delta=?$ $\sharp$ $\delta=\epsilon\cdot\sqrt{x_0}$
\end{enumerate}
\item[$\Vdash$] $\{\text{show the two conjuncts separately, use generalization rule for second conjunct}\}$
\begin{itemize}
\item Show that $|f(x)-f(x_0)|<\epsilon$, when
\begin{enumerate}[resume]
\item\label{item:xnonneg} $x\geq 0$, and
\item $|x-x_0|<\delta$
\end{enumerate}
\item[$\Vdash$] $|f(x)-f(x_0)|$
\item[$=$] $\{$assumption \ref{item:range}, $f(x)$ and $f(x_0)$ are defined by assumptions \ref{item:xnonneg} and \ref{item:range}$\}$
\item[] $|\sqrt{x}-\sqrt{x_0}|$
\item[$=$] $\{$extend with conjugate value for $\sqrt{x}-\sqrt{x_0}$, i.e., with $\sqrt{x}+\sqrt{x_0}>0$$\}$
\item[] $\dfrac{|x-x_0|}{\sqrt{x}+\sqrt{x_0}}$
\item[$\leq$] $\{$make divisor smaller, $\sqrt{x}\geq 0$ $\}$
\end{itemize}
\end{itemize}
\end{itemize}
\end{itemize}
\end{document}
