A connected subspace if connected in a subspace that contains it? I think I saw somewhere before that if $A \subset Y \subset X$, $A$ is connected in $X$ if and only if $A$ is connected in $Y$ ($Y$ with the subspace topology), but do not recall the proof.
So I tried to prove it. Here's my progress.
$\implies$
Suppose $A$ is connected in $X$. Then, there exists no separation $U$ and $V$ of $A$ (i.e, there exists no pair of non-empty disjoint sets $U$ and $V$ such that $\overline U \cap V = \varnothing$, $U \cap \overline V = \varnothing$, and $U \cup V = A$). Denote $cl_Y()$ as a closure in $Y$; note that $\overline U$ meant and will be used to denote the closure of $U$ in $X$ as usual. Then, if there exists a separation of $A$ in $Y$, then there exists some non-empty disjoint sets $U'$ and $V'$ such that $cl_Y (U') \cap V' = \overline U' \cap Y \cap V' = \varnothing$, $U' \cap cl_Y(V') = U' \cap \overline V' \cap Y = \varnothing$, and $U' \cup V' = A$. Since $U', V' \subset A \subset Y$, $U' \cap \overline V' \cap Y = U' \cap \overline V'$ and $\overline U' \cap V' \cap Y = \overline U' \cap V' = \varnothing$, so $U'$ and $V'$ would be a separation of $A$ in $X$. Thus, by contradiction, $A$ must be connected in $Y$.
$\impliedby$
Suppose $A$ is connected in $Y$. As above, there exists no disjoint non-empty sets $U$ and $V$ such that $\overline U \cap Y \cap V = \varnothing$, $U \cap Y \cap \overline V = \varnothing$ and $U \cup V = A$. If there exists a separation $U'$ and $V'$ of $A$ in $X$, both sets are non-empty, $\overline U' \cap V' = \varnothing$, $U' \cap \overline V' = \varnothing$, and $U' \cup V' = A$. Then, $cl_Y(U') \cap V' = \overline U' \cap Y \cap V' \subset \overline U' \cap V' = \varnothing$ and $cl_Y(V') \cap U' \subset U' \cap \overline V' = \varnothing$, $U'$ and $V'$ would be a separation of $A$ in $Y$, which is a contradiction.
I thought I was stuck, but now that I write about it, I think above might be correct. Can someone verify?
 A: I prefer the more "intrinsic" definition of connectedness that does not refer to a larger space.
A space $X$ is called disconnected if we can write $X=A \cup B$ with $A \cap B = \emptyset$, $A \neq \emptyset$, $B \neq \emptyset$ and such that one of the following equivalent conditions are fulfilled :

*

*$A$ and $B$ are separated (so $\overline{A} \cap B =\emptyset = A \cap \overline{B}$).

*$A$ and $B$ are both closed.

*$A$ and $B$ are both open.

(1 implies 2 as $A \subseteq \overline{A} \subseteq X\setminus B = A$  and symmetrically for $B$ etc., 2 implies 3 as $A$ and $B$ are each other's complement, and 3 implies 1 because $O \cap A = \emptyset$ implies $O \cap \overline{A}  =\emptyset$ for open sets $O$.)
Finally, $X$ is defined to be connected when it's not disconnected.
A ssubset $A$ of a space $X$ is connected when it is connected (in the above sense) in its subspace topology.
And when we have $A \subseteq Y \subseteq X$ then it is standard that the subspace topology that $A$ inherits from $Y$ (in its subspace topology) is the same as the subspace topology it gets directly from $X$ itself. ($(O \cap Y) \cap A = O \cap A$, is what the proof comes down to.) So there is nothing to proev really, if you follow this definition.
Note that this definition is equivalent to the Munkres' one you said you were using in the comments.
A: The easiest way to do this is by picking the right definition of "connected". First, define a space $X$ to be connected iff for every open $U, V$, $U \cap V = \emptyset$, $U \cup V = X$, either $U = X$ or $V = X$. Then define a subset $S \subseteq X$ to be connected iff $S$ is a connected space under the subspace topology.
Then if $A$ is a subspace of both $B$ and $C$, then $A$ is connected iff it is a connected subset of $B$ iff it is a connected subset of $C$.
The definition of "connected set" you appear to be using is that $A \subseteq X$ is connected iff for every $U, V$ open, $U \cap V \cap A = \emptyset$, $A \subseteq U \cup V$, either $A \subseteq V$ or $A \subseteq U$. This is clearly equivalent to $A$ being connected under the subspace topology by the definition of that topology.
