They are equivalent.
Let $X$ be disconnected as in the first sense.
Then there are (nonempty) open sets $A$, $B$ in $X$ such that $X = A \cup B$ and $A \cap B = \emptyset$, that is $X\setminus A = B$.
Now since $A$ is open in $X$, the set $B$ is the complement of an open set, so $B$ is closed and $\overline B = B$.
Therefore, $A \cap B = \emptyset$ becomes $A \cap \overline B = \emptyset$. Do the same for $A$ and $\overline A \cap B$.
So the space is the union of two separated sets and therefore not connected.
Let $X$ be not connected as in the second sense.
Then there are (nonempty) separated sets $A$, $B$ in $X$ such that $X = A \cup B$.
Since $A \cap \overline B = \emptyset$, you have $A = X\setminus \overline B$, so $A$ is open in $X$. Do the same for $B$.
From $A \cap \overline B = \emptyset$ you also get $A \cap B = \emptyset$, since $B \subseteq \overline B$.
Therefore, the space is the union of nonempty open sets, so the space is disconnected.
Now, the definition given by Rudin mentions subsets, but extending the first definition by calling a subset $E \subseteq X$ connected if $E$ is connected with respect to the subspace topology, and checking that $E \subseteq X$ is disconnected if and only if $E \subseteq E$ is disconnected with respect to the subspace topology, you see, that both notions are really the same.
