Your definition is that a subset of $\Bbb R$ is closed if it contains all of its limit points, so in order to show that $A\cup L$ is closed, you must show that if $x$ is a limit point of $A\cup L$, then $x\in A\cup L$.
Suppose that $x$ is a limit point of $A\cup L$. Then for each $\epsilon>0$, $(x-\epsilon,x+\epsilon)$ contains a point of $A\cup L$ different from $x$. If every one of those open intervals contains a point of $A$ different from $x$, then $x$ is a limit point of $A$, so $x\in L\subseteq A\cup L$, and we’re done. Suppose, hoping to get a contradiction, that $x$ is not a limit point of $A$. Then there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)$ is either empty or just $\{x\}$. On the other hand, $x$ is a limit point of $A\cup L$, so $(x-\epsilon,x+\epsilon)$ contains some $y\in A\cup L$ with $y\ne x$. Clearly this means that $y\in L\setminus A$. Now let $\delta=\min\{y-(x-\epsilon),(x+\epsilon)-y\}$.
Show that $(y-\delta,y+\delta)\subseteq(x-\epsilon,x+\epsilon)$.
Your corrected proof is correct. Here's a shorter and simpler way to prove the second direction: Let X be the topological space where $A\subseteq X$.Recall that the closure of a set $\bar A$ =$A\cup A'$ where A'= { x | x is an accumulation point of $A\subseteq X$.Recall also that $\bar A$ is a closed set because it is the intersection of all the closed subsets of X that contain A as a subset and the intersection of any number of closed sets is closed. So assume $A\subseteq X$ contains all it's accumulation points. Then $A'\subseteq A$.Then $A\cup A'\subseteq A$.But $A\subseteq A\cup A'$. Therefore, $A = A\cup A'$=$\bar A$ and therefore A is closed! Q.E.D.
In closing, here's one for you to mull your mind over: Consider $A\subseteq X$ where X is a topological space. Assume A has no limit points. Then is it closed? Why or why not?
Best Answer
Let $X$ be a topological space and $S$ a closed subset of $X$.
If $x$ is a limit point of $S$, then each open set $U$ with $x \in U$ has the property that $ U \cap (S \setminus \{x\}) \ne \emptyset.$
Now suppose that $x \notin S$. Then $U:= X \setminus S $ is open and $x \in U.$
Then we have $ U \cap (S \setminus \{x\}) \ne \emptyset$ , on the other hand we have $ U \cap (S \setminus \{x\}) = \emptyset$, a contradiction.
So: $x \in S.$