A simple graph $G$ is connected iff for every partition of the vertices into two non-empty sets $X$ and $Y$, there is a vertex $x\in X$ and a vertex $y\in Y$ such that $xy$ is an edge of $G$.
My proof:
$(\Rightarrow: )$ Since $G$ is connected, then every pair of vertices $v,w$ there is a path in $G$ from $v$ to $w$. We partition the vertices into two non-empty sets $X$ and $Y$, pick two arbitrary vertices $x\in X$ and $y\in Y$, then there is a path from $x$ to $y$. The path contains at least $1$ edge. We pick the edge where the endpoints are one in $X$ and other in $Y$ (but I am really not sure how to explicitly show this). Let the endpoints be $x'\in X$ and $y'\in Y$ and hence $x'y'$ is an edge in $G$.
$(\Leftarrow: )$ We arbitrarily partition the vertices of $G$ into two non-empty sets $X$ and $Y$, let the vertices $x\in X$ and $y\in Y$ be such that $xy$ is an edge in $G$. Since the partition is arbitrary, there always exists a path from any $x'\in X$ to $y'\in Y$ (I am also not so sure how to write this part rigorously). Hence we can conclude that $G$ is connected.
Is my proof correct? And how can I fill in some missing parts in the proof?
The thing that I doubt is that I did not use the fact that $G$ is a simple graph. How should I include that in my proof? I am not sure whether the fact that $G$ is simple is important in the statement.
Could somebody please give some help and clarification on it? And some advice on how to improve the proof please? Thanks!
The ($\Rightarrow$:) part is almost complete, except where you are not sure. We pick $x \in X$ and $y \in Y$, and we know that there is a path between them. Let the vertices in this path be $x, a_1, a_2, \dotsc, a_n, y$. The first vertex is in $X$, and the last vertex is in $Y$; all vertices are either in $X$ or $Y$ and never in both. Consequently, there will be two adjacent vertices in the path where one is in $X$ and the other is in $Y$ (if this never happened, how do you get from $X$ to $Y$?). Let these vertices be $x'$ and $y'$, respectively. Since they are adjacent in the path, there is an edge between them and the proof is complete.
For the ($\Leftarrow$:) part, a different approach is required (in my opinion). If $G$ has no vertices or only one vertex, then the proof is trivial. If there is more than one, let $X$ contain only one vertex $x$ and let $Y$ contain the rest. By our assumption, for some $y \in Y$, the edge $xy$ exists.
Now include $y$ in $X$ and remove it from $Y$. Note that $X$ is a connected graph because a path exists between any two of its vertices. Repeat the above procedure: for some $x_1 \in X$ and some $y_1 \in Y$, the edge $x_1y_1$ exists. Now include $y_1$ in $X$ and remove it from $Y$. Again, note that $X$ is connected because for any two vertices in $X \setminus y_1$, a path exists between them; and a path exists between one vertex in that set and $y_1$, so they are all connected. Repeat this procedure until all the vertices are contained in $X$. Now $G$ itself is connected, and the proof is complete.
(I, too, cannot see why $G$ must be simple, so perhaps this is a more general property than what you have been asked to prove.)