Let $(M, d)$ be a metric space and $A\subset M$. Show that $U\subset A$ is open The entire question is as follows:
Let $(M,d)$ be a metric space and $A\subset M$.  Consider the metric space $(A,d)$.
Show that a set $U\subset A$ is open in $(A,d)$ if and only if there exists an open set $V$ in $(M,d)$ such that $U = A\cap V$.
I'm struggling on this particular problem and in topology proofs in general.  I'm having a hard time applying the definitions to solve problems/develop proofs.
I'm thinking this has something to do with connected sets. Intuitively, I can tell that, with $V$ open, it would follow that $U$ is open.  Should I proceed by showing the complement of $U$ is closed?  Here's what I have tried:
Let $x_0 \in U$.  Since $V$ is open, $V = int(V)$ and $x_0 \in int(V)$.  I then need to show that $x_0$ is in int(A), but I'm not sure how to do that.
If I consider connectedness, I recall the definition of separate being two open subsets $U, V$ that satisfy:
(a) $U \cap V \cap A = \emptyset$
(b) $A \cap U \neq \emptyset$
(c) $A \cap V \neq \emptyset$
(d) $A \subset U \cup V$
It is my understanding that all of these must be true to be disconnected.  So, since $U \cap V \cap A \neq \emptyset$, A is connected, but how do I prove that U is open here?  The grader has shown in the past that citing the definition is not sufficient.  You must show work.  How do I go from this definition, which appears pretty straightforward, to a proof? 
 A: I'm not sure why you thought in connected sets, but the solution to that problem is actually much simpler: Let $d$ be the metric given in $M$. Given $X\subseteq M$, $x\in X$ and $\varepsilon>0$, let's write the open ball $B^X_\varepsilon(x)=\left\{y\in X:d(x,y)<\varepsilon\right\}$. Notice that, if $x\in Y\subseteq X\subseteq M$ and $\varepsilon>0$, then $B^Y_\varepsilon(x)=B^X_\varepsilon(x)\cap Y$.
$(\Leftarrow)$ Suppose that $U=A\cap V$ for some $V$ open in $M$. Let's show that $U$ is open in $(A,d)$.
Let $x_0\in U$. In particular, $x_0\in V$, so there exists $\varepsilon>0$, such that $B^M_\varepsilon(x_0)\subseteq V$. Notice that $B^A_\varepsilon(x_0)=B^M_\varepsilon(x_0)\cap A\subseteq V\cap A=U$. Therefore, $U$ is open in $A$.
$(\Rightarrow)$ Suppose that $U$ is open in $A$. Let's find an $V$ open in $M$ such that $U=A\cap V$.
Let $\mathscr{V}=\left\{W\subseteq U: W\text{ is open in }M\text{ and }W\cap A\subseteq U\right\}$ and $V=\bigcup\mathscr{V}:=\bigcup_{W\in\mathscr{V}}W$. Since $V$ is an union of open subsets of $M$, $V$ is open in $M$ itself. Let's show that $V\cap A=U$. By the definition of $\mathscr{V}$, it's clear that $V\cap A\subseteq U$.
For the oposite inclusion, let $x_0\in U$. Then there exists $\varepsilon>0$ such that $B^A_\varepsilon(x_0)\subseteq U$. Then $B^M_\varepsilon(x_0)$ is an open set in $M$ and $B^M_\varepsilon(x_0)\cap A=B^A_\varepsilon(x_0)\subseteq U$, so $B^M_\varepsilon(x_0)\in\mathscr{V}$, thus $B^M_\varepsilon(x_0)\subseteq V$. In particular, $x_0\in B^A_\varepsilon(x_0)=B^M_\varepsilon(x_0)\cap A\subseteq V\cap A$.
Therefore, $V\cap A=U$, as we wanted.
A: Notation: 
$$B(x,r)=\{y\in M | d(x,y) < r_x\}
$$



*

*Assume that $U = A\cap V, V$ open set of $M$, let $x\in U$. As $x\in V$, there is a $r>0$ such as $B(x,r)\subset V$; $$
\{y\in A | d(x,y) < r\} = A\cap B(x,r)\subset A\cap V=U
$$
hence $U$ is a neighborhood of $x$ in $A$, and $U$ is open in $A$.

*If $U$ is an open set of $A$: for every $x$ there is $r_x$ such as 
$$\{y\in A | d(x,y) < r_x\}\subset U.
$$
Actually, there is even $U = \cup_{x\in U} \{y\in A | d(x,y) < r_x\}$.
Now consider $V = \cup B(x,r_x)$. This is an open set of $M$ (union of open sets) such as $$U = \bigcup_{x\in U} \{y\in A | d(x,y) < r_x\}
= \bigcup_{x\in U} A\cap B(x,r_x) = A\cap\bigcup_{x\in U}B(x,r_x) = A\cap V.
$$
A: This has nothing to do with connectedness.
Use the fact that a set is open if and only it's it's a union of open balls.
If $U\subseteq A$ is open in $A$, then $U=\bigcup_{i\in I} B_i^A$ and set $V=\bigcup_{i\in I}B_i^M$ of the correspondent balls in M.
Similarly the other direction.
