In "Foundation of Topology" by C.W Patty, given
$$\begin{align}A :=& \ \{(x,y)\in \mathbb R^2:y=0\},\\ X:=& \ A\cup\{(0,1)\},\end{align}$$
the topology on $X$ is defined as
$$\tau:= \ \{U\in \mathscr P(X):U=\varnothing \vee (0,1)\in U\}.$$
I am able to prove the given collection is a topology on $X$ without knowing how $U$ looks like .
My attempt of a proof:-
1) $\varnothing \in \tau$[$\because$ from definition] $X\in \tau [\because X=A\cup\{(0,1)\}$
2) $(0,1)\in U_\alpha,\alpha \in \Gamma$, $(0,1)\in \cup_{\alpha}U_{\alpha}\implies \cup_{\alpha}U_{\alpha}\in \tau$
3) $(0,1)\in U_\alpha,\alpha \in \Lambda$,$|\Lambda|=$finite, $(0,1)\in \cap_{\alpha}U_{\alpha}\implies \cap_{\alpha}U_{\alpha}\in \tau$.
I don't know how would the elements look like?
How $\{(0,1)\}$ is a dense subset in $X$?
Please help me to visualize the open set in $\tau$.
An open set is the empty set $\varnothing$ or any subset of the $x$-axis together with the point $(0,1)$. You can visualize this to the extent that you can visualize arbitrary subsets of $\mathbb{R}$.
A closed set is the whole space $X$ or any subset of the $x$-axis (as the complement of an open set, a closed set does not contain $(0,1)$, unless it is $X$).
Why is $\{(0,1)\}$ dense? Well, the closure of $\{(0,1)\}$ is the intersection of all closed sets containing $(0,1)$, and there's only one of these, namely $X$.
Your proof that $\tau$ is a topology looks mostly fine, though you've forgotten to handle the cases when some $U_\alpha$ are empty in (2) and (3). Also, it's hard to read because you're writing things using symbols that would be clearer written out in words. For example, for (3), I would write: