If $A$ and $B$ are compact sets in $M$, show that $A\cap B$ is compact If $A$ and $B$ are compact sets in $M$, show that $A\cap B$ is compact.
Could someone please check my proof for this problem? My working definition of "compact" is

$P$ is compact iff every sequence in $P$ has a subsequence that
  converges to a point in $P$.

My attempt:
Let $(x_n)$ be a sequence in $A\cap B$.
$x_n\in A\cap B$
$\implies x_n\in A$ and $x_n\in B$
So, $(x_n)$ is a sequence in $A$. Since $A$ is compact, $(x_n)$ has a subsequence $(x_{n_i})$ that converges to a point $y\in A$. 
$(x_n)$ is a sequence in $B$. Since $B$ is compact, $(x_n)$ has a subsequence $(x_{m_i})$ that converges to a point $z\in B$.
This is where I'm stuck. How do I proceed? I considered taking the common terms of the 2 subsequences but there need not be any.
 A: I assume the space involved are Hausdorff since you use subsequences. The idea is to consider $x_{n_i}$as a subsequence of $B$ since $x_{n_i}$ is also in $B$ which  has a subsequence  $x_{n_{i_j}}$ which converges and the limit is also $y$.
A: You want to show that every open cover of $A\cap B$ has a finite sub-cover.
Take an open cover of $A\cap B$ and augment it with the complement of $A\cap B$
Note that this complement is open and it covers everything which in not in $A\cap B$
This  augmented covering  will cover both $A$ and $B$ 
Thus it has a finite sub-cover for $A$ and  a finite sub-cover for $B$
The union of these two covering is a finite open covering for $A\cap B $
Thus $A\cap B $ is compact.   
A: You take the sequence $(x_n) \subset A \cap B$, then $(x_n) \subset A$, so there exists $(x_{n_i})$ subsequence that converges to a $y \in A$
Then $(x_{n_i}) \subset A \cap B$, so it has a subsequence $(x_{n_{i_j}}) \subset B$ that conveges a point $z \in B$
Then $y = z$ 
Sorry for my bad english
