let $A$ be a finite open subset of a metric space $M$ prove that every point in $A$ is an isolated point of $M$ Is the following approach correct?

Since $A$ is a open subset (finite) of the metric space $M$, we can use the following theorem(s) which states:

Theorem I. A subset of a metric space is open i.f.f it contains, along with any point $x$, some neighborhood of $x$
Theorem II. A subset of a metric space is open i.f.f it is expressible as a union of open balls.

Therefore, since we already have the information that $A$ is a open subset, it can be
I. Expressible as a union of open balls [not assumed to be disjoint]
II. Every point $x$ in $A$ has a neighborhood.

''Definition'' picked from WIKIPEDIA In mathematics, a point $x$ is called an isolated point of a subset $S$ (in a topological space $X$) if $x$ is an element of $S$ but there exists a neighborhood of $x$ which does not contain any other points of $S$.

&quot;Therefore, every $x$ in A is an isolated point of $M$ since every $x$ according to above has a neighborhood (balls)...&quot; [not conclusive enough]
From here, I do struggle a bit concluding.
Does anyone have any suggestions?
You can write in a simpler way :

Let $p \in A$. Because $A$ is finite, $A \setminus \lbrace p \rbrace$ is finite so it's closed in $M$. So $M \setminus \left( A \setminus \lbrace p \rbrace \right)$ is open. Now note that 
$$\lbrace p \rbrace = A \cap \left( M \setminus \left( A \setminus \lbrace p \rbrace \right) \right)$$

is the intersection of two open sets, so it is open. So $p$ is isolated in $M$.
Let X=$\{x_1, ..., x_n\}$ be the points of your open subset. Since your space is metric, it is separated and so $\forall i != j , \exists B_{i,j} , B_{j,i}$ two open balls such that $i \in B_{i,j}, j \in B_{j,i}, B_{j,i} \cap B_{j,i} = \emptyset$, then define $\forall i, B_i = \bigcap_{j=1}^n B_{i,j}$, then $\forall i != j , x_j \not \in B_i$, and as a result $B_i \cap X - \{x_i\} = \emptyset$ and every point of X is isolated.