Proving that a set $F \subset A$ is closed in $(A,d)$ if and only if $F = A \cap C$, where $C$ is closed in $(M,d)$ I need to proof that, given a metric space $(M,d)$ and a subset $A \subset M$ that:

A set $F \subset A$ is closed in $(A,d)$ if and only if $F = A \cap C$, where $C$ is closed in $(M,d)$

I have tried alot of things, but I can't seem to make it work. Can anyone help me out?
Thanks,
K. Kamal.
 A: amsmath's comment is the "best" way to do it if you have seen the definition of relative openness. However, since you are working with metric spaces, I'm not convinced that you have. Let's see if we can do it with sequences.
We use the definition that if $E$ is a subset of a metric space $(X,d)$, then $\newcommand{\cl}{\operatorname{cl}}x\in\cl(E)$ iff there exists a sequence $(x_n)$ in $E$ which converges to $x$.
We will use $\cl_A$ to denote the closure of a set in $(A,d)$ (which should technically be $(A,d|_{A\times A})$), and $\cl_M$ to denote the closure of a set in $(M,d)$.
Suppose $F=A\cap C$, where $C$ is closed in $(M,d)$. In order to show that $F$ is closed in $(A,d)$, we must show that $F\supseteq \cl_A(F)$. That is, we need to prove that if $x\in A$ such that there exists a sequence $(x_n)$ in $F$ which converges to $x$ in $(A,d)$, then $x\in F$.
Observe that $(x_n)$ also converges to $x$ in $(M,d)$.
Since $(x_n)$ is in $F=A\cap C$ and $C$ is closed in $(M,d)$, then its limit point must be in $\cl_M(C)=C$. Thus $x\in A\cap C=F$, as desired.
Conversely, assume $F$ is closed in $(A,d)$. We wish to find a closed set $C$ in $(M,d)$ such that $F=A\cap C$. Take $C:=\cl_M(F)$. Then clearly $F\subseteq A\cap C$.
For the other inclusion, fix $x\in A\cap C$. Then $x\in A$ and $x\in C= \cl_M(F)$, so there is a sequence $(x_n)$ in $F$ which converges to $x$ in $(M,d)$. Since $(x_n)$ is in $F\subseteq A$ and $x\in A$, we have that $(x_n)$ converges to $x$ in $(A,d)$. Therefore $x\in \cl_A(F)=F$.
