I am aware of a quite similar question, exactly at the same 19A question of Willard's General Topology, but mine is for the first item:
- This is a valid assignment of neighbourhoods in $Y^*$.
As a reminder, the construction, by Willard, of the Alexandroff extension of any topological space $Y$, is:
$$Y^* = Y \cup \{p\}$$
$$\forall x \in Y, \mathscr{B}_{Y^*,x} = \mathscr{B}_{Y,x}$$
$$\mathscr{B}_{Y^*,p} = \{\{p\} \cup (Y \setminus L), L \subset Y \wedge L ~\mathrm{compact}\}$$
(Where $\mathscr{B}_{A,x}$ is the neighbourhood base at $x$ in the topological space $A$.)
Also, any topology built from neighbourhoods need to ensure that necessary conditions are met (else that could fail to define a topology).
One of them is (called V-c at section 4.5 page 33 of Willard):
if $V \in \mathscr{B}_x$, there is some $V_0 \in \mathscr{B}_x$ such that if $y \in V_0$, then there is some $W \in \mathscr{B}_y$ with $W \subset V$
I fail to prove this property if $L$ is only compact and not closed. I cannot prove that it is closed, so that should be part of the hypotheses.
So, even if seemingly the condition that $L$ is closed would not be required for other items of the exercise, however if it is required to ensure that a topology has been defined on $Y^*$, then other question would loose their meaning without it...
However, me being a noob, and very much likely to make gross mistakes, I would be grateful of some validation / check...
Your doubts are justified.
Willard defines neigborhood bases for all points of $Y^*$ and the implicit claim is that their union $\mathscr{B}$ is a base for some topology. By Theorem 5.3 we must show that for $B_1, B_2 \in \mathscr{B}$ and $y \in B_1 \cap B_2$ there exists $B \in \mathscr{B}$ such that $y \in B \subset B_1 \cap B_2$. This requires in particular that for each $B' = \{p\} \cup (Y \setminus L)\in \mathscr{B}_{Y^*,p}$ and each $y \in B' \cap Y$ there exists $B \in \mathscr{B}$ such that $y \in B \subset B' \cap Y$. Clearly we must have $B \in \mathscr{B}_{Y,y}$. But if $L$ is not closed in $Y$, then this is not true since $B' \cap Y = Y \setminus L$ is not open in $Y$ which means that some $y \in Y \setminus L$ does not admit $B \in \mathscr{B}_{Y,y}$ with $B \subset Y \setminus L$.
In fact, the standard definition of the Alexandroff extension is to require that $L$ is compact and closed. See https://en.wikipedia.org/wiki/Alexandroff_extension.
One could of course take Willard's system as a subbase for some topology. As an example let us consider any space $Y$ with more than one point and having the trivial topology. Then all $L \subset Y$ are compact, but non-closed for $L \ne \emptyset, Y$. So let $y \in Y$ and $L = Y \setminus \{y\}$. The topology generated by Willard's system must have the property that $y$ has an open neighborhood $B$ in $Y^*$ such that $y \in B \subset (Y^*\setminus L) \cap Y = \{y\}$. If we want that $Y$ is a subspace of $Y^*$ (as Willard requires in his exercise), then $\{y\}$ must be open in $Y$ which is not true. This shows that the subbase approach does not make sense.