here there are 2 definitions of locally closed sets:
$A$ is locally closed subset of $X$ if:
a) every element in $A$ has a neighborhood $V$ in $X$ such that $A\cap V$ is closed in $V$.
b) $A$ is open in its closure (in $X$)
why a) and b) are equivalent?
For me the definition is the following:
$c)$ $A$ is equal to $U\cap F$ where $U$ is open in $X$ and $F$ is closed in $X$.
Let us prove that this is equivalent to your definitions, that I write again here:
$a)$ For each $x\in A$ there is a neighbourhood $V$ of $x$ in $X$ such that $A\cap V$ is closed in $V$.
$b)$ $A$ is open in its closure $\overline{A}$ in $X$.
The proof is made in the following steps:
$c)\Rightarrow a)$ Just choose $V=U$, which is a neighbourhood of any point of $A$.
$a)\Rightarrow b)$ For each point $x\in A$ we use the given neigbourhood $V$ such that $A\cap V$ is closed in $V$. We can replace $V$ with an open neighbourhood $U\subset V$ (since $A\cap U=(A\cap V)\cap U$ is closed in $U$), and assume that $V$ is open.
The fact that $V$ is open implies that the closure of $A\cap V$ in $V$ is equal to $\overline{A}\cap V$. Indeed, $A=(A\cap V)\cup (A\cap (X\setminus V))$ so $\overline{A}\subset \overline{A\cap V}\cup \overline{A\cap (X\setminus V)}$. As $V$ is open we have $\overline{A\cap (X\setminus V)}\subset X\setminus V$, hence $V\cap \overline{A\cap (X\setminus V)}=\emptyset$, so we get $$\overline{A}\cap V\subset \overline{A\cap V} \cap V.$$ The other implication being clear, and using the hypothesis that $A\cap V$ is closed in $V$ we get$$\overline{A}\cap V=\overline{A\cap V} \cap V=A\cap V\subset A.$$
Hence, $\overline{A}\cap V$ is a neighbourhood of $x$ in $\overline{A}$ contained in $A$. Doing this for all points of $A$, this shows that $A$ is open in $\overline{A}$.
$b)\Rightarrow c)$ Since $A$ is open in $F=\overline{A}$, there exists an open subset $U$ of $X$ such that $\overline{A}\cap U=A=F\cap U$.
b) $\Rightarrow$ a)
Let $A$ be open in $\overline{A}$. Then $A=\overline{A}\cap U$ for some $U$ open in $X$ and $U$ serves as neighborhood for every element of $A$. Intersection $A\cap U$ equals $A=\overline{A}\cap U$ which is a subset of $U$ closed in $U$.
a) $\Rightarrow$ b)
Let $N$ be a neighborhood of $a\in A$ such that $N\cap A$ is closed in $N$. Its closure in $N$ is $N\cap\overline{A}$ so actually we have $N\cap A=N\cap\overline{A}$. That shows that $N\cap A$ is a neighborhood of $a$ in the subtopology on $\overline{A}$, and it is contained in $A\subset\overline{A}$. For any $a\in A$ such a neighborhood exists, so $A$ is open in $\overline{A}$.
Just for the sake of completeness, I will state the complete list of equivalences:
Proposition. Let $X$ be a space and $A\subset X$ be a subspace. The following are equivalent:
(1) There is $U\subset X$ open such that $A\subset U$ and $A$ is closed in $U$.
(1') $A$ is the image of the composition of a closed immersion followed by an open immersion.
(2) There is $F\subset X$ closed such that $A\subset F$ and $A$ is open in $F$.
(2') $A$ is the image of the composition of an open immersion followed by a closed immersion.
(3) $A=U\cap F$, for some $U\subset X$ open and $F\subset X$ closed subsets.
(4) For all $x\in A$ there is an open neighborhood $U\subset X$ of $x$ such that $A\cap U$ is closed in $U$.
(5) For all $x\in A$ there is a neighborhood $U\subset X$ of $x$ such that $A\cap U$ is closed in $U$.
(6) $A$ is open in $\overline{A}$.
(7) There is $U\subset X$ open such that $A\subset U$, $A$ is closed in $U$, and $U\cup \overline{A}=X$.
If any of these equivalent conditions happens, we say that $A$ is locally closed in $X$. Moreover, if $A$ is locally closed in $X$ and $U\subset X$ is an in (7), then $U=(X\setminus\overline{A})\cup A$ and $U$ is the maximum in open subsets $V$ of $X$ that contain $A$ and such that $A$ is closed in $V$.
The proof of the equivalences (1)$\Leftrightarrow$(1'), (2)$\Leftrightarrow$(2'), and (1)$\Leftrightarrow$(2)$\Leftrightarrow$(3) is immediate. The rest of the equivalences that do not involve (7) were already proven by Jérémy Blanc on his answer.
(6)$\Rightarrow$(7). Since $\overline{A}\setminus A$ is closed in $X$, the set $U=X\setminus(\overline{A}\setminus A)=(X\setminus\overline{A})\cup A$ is open in $X$. It is clear that $\overline{A}\cup U=X$. The closure of $A$ in $U$ equals $\overline{A}\cap U=A$.
(7)$\Rightarrow$(6). We have $X\setminus\overline{A}=U\setminus\overline{A}=U\setminus A$, where the last equality follows since $\overline{A}\cap U=A$, for $A$ is closed in $U$. Hence $U=(X\setminus\overline{A})\cup A=X\setminus(\overline{A}\setminus A)$ is open, so $\overline{A}\setminus A$ is closed. Thus, $A$ is open in $\overline{A}$.
Finally, to see the maximality of $U=(X\setminus\overline{A})\cup A$, let $V$ be an open subset of $X$ containing $A$ and such that $A$ is closed in $V$. Let $x\in V$. If $x\in A$, then $x\in U$. If $x\notin A$, then $x\in V\setminus A=V\setminus\overline{A}\subset X\setminus\overline{A}\subset U$ (the first equality follows from the fact that $A$ is closed in $V$). Hence $V\subset U$.
