There is a conclusion :Suppose $M$ is a connected topological manifold,then $M$ is path-connected.
We can define $S_p=\{q\in M:$there is a path from $p$ to $q\}$.Obviously,$S_p$ is nonempty since $p\in S_p$,If we prove that $S_p$ and $M\setminus S_p$ is open,then we can conclude that $M=S_p$ by connectness of $M$.
Suppose $q\in S_p$,then there is a coordinate chart $(U,\phi)$ around $q$ and $\phi(U)$ is open in $R^n$ for some $n$,how to prove $S_p$ is open by using the above fact?
If $q \in S_p$ then, as you said, there is a coordinate chart $(U,\phi)$ such that $U \ni q$ and $\phi(U)$ is -without loss of generality- path connected (indeed you can consider an open ball $B$ -which is path connected- of $R^n$ containing $\phi(q)$ and contained in $\phi(U)$ and then restrict your chart to $\phi^{-1}(B)\subset U$).
Hence, if $q' \in U$, you can consider a path in $\phi(U)$ connecting $\phi(q)$ and $\phi(q')$. Then you consider the inverse image of this path through $\phi$ in order to get a path in $U$ connecting $q$ and $q'$ ($\phi$ is bijective). Hence $q'\in S_p$ because you already know that there is a path connecting $p$ and $q$. And this proves that $U \subset S_p$. As $U$ is open and $q \in U$, we are done.
More briefly, as $\phi(U)$ can be choosen to be path connected and as $\phi$ is an homeomorphism, $U$ can be choosen to be path connected.
