connected manifolds are path connected 
prove every connected manifold is path connected manifold . 
my thought:
connected space : Let $ X$  be a topological space. A separation of $ X $ is a pair $U, V$ of disjoint 
  nonempty open subsets of $ X $ whose union is $X$. The space $ X $ is said to be connected 
  if there does not exist a separation of  $X$ .
Components: Given  $X$ , define an equivalence relation on $X$ by setting x~y if there 
  is a connected subspace of $X$  containing both $ x$  and $ y$ . The equivalence classes are 
  called the components (or the "connected components") of $ X$. 
Path Component: I define another equivalence relation on the space $ X$  by defining x ~ у 
  if there is a path in $ X$  from $ x$  to $ y$ . The equivalence classes are called the path components of $ X$ . 
Theorem  : The path components of $ X$  are path connected disjoint subspaces 
  of $X$  whose union is $X$ , such that each nonempty path connected subspace of $X$  intersects only one of them. 
thank you so much

 A: path connected sets are also connected, conversely is not always true (there is a famous counterexample which shows this fact). However a necessary condition To make the converse of the previous  statement possible is that the connected sets are required to be locally Euclidean. So a topological manifold, by definition is also a locally Euclidean space, if you can prove the previous statement you can get the answer. Hope it helps
A: Let $x\in X$. Consider $U:=\{y\in X\mathop{|}\text{there is a path from $x$ to $y$}\}$. So $U$ is nonempty: $x\in U$. Claim: $U$ and $U^c:=X\smallsetminus U$ both are open. To prove this use the fact that given any point $z\in X$ there is a neighbourhood of $z$ which is homeomorphic to an open ball of $\mathbb{R}^n$ for some $n$. The homeomorphic image of a path connected set is path connected, so $U$ and $U^c$ both are open but $U \cup U^c = X$ which implies that $U=X$ since $U$ is nonempty.
A: Given our manifold M is connected, let $x, y \in M$ be arbitrary point.
Since for any $x$ there is an open set $U_x \cong \mathbb{R}^n$ for which  $x$ is contained, let $C_x$ denote the path component of $x$. If $y$ is contained then we are done. However, if not then as then let $\bigcup_{x \in C_x}U_x =C_x $ which is open. 
$M$ is the union of path connected components, by previous reasoning all path connected components are open. Thus this implies that $M$ is disconnected  contradicting our hypothesis $y \notin M$. Therefore, $M$ is path connected.
Conversely, path connectednesss implies connectedness as required.
Proving that Manifold is connected if and only it is path connected.
