Is the graph of a continuous function from a connected space also connected? Let X be connected and $f:X \rightarrow Y$ continuous. Is the graph $G$ of this function connected?
My thought is yes, which seems pretty intuitively clear. 
For contradiction, assume $G$ is not connected and $G= U \cup V$. I pick $(x,y)$ in $G$, so WLOG assume $(x,y)$ is in $U$. But since $U=A \times B$ and $V=C \times D$, the disconnectedness of U and V implies B and D are disjoint. This implies the range of f is disconnected which implies X is disconnected, which is a contradiction.
I don't think this is right because I really don't need to use the contuity of f at all. And it ended up not mattering if I chose an (x,y).
 A: If $f :X \to Y$ is a function, $X$ connected, then graph of $f$ is merely the subspace of $X \times Y$ represented by $G_f = \{(x, f(x)): x\in X\}$. Thus, it's the image of the function $f : X \to X \times Y$ given by $x \mapsto (x, f(x))$, and since continuous image of a connected space is connected, graph of $f$ is connected.
One a different note: the converse, however, is false. Consider the map $f : [0, 1] \to [0, 1]$ given by $f(x) = \sin(1/x)$ for $x \neq 0$ and $0$ otherwise. Graph of this is the subspace of $[0, 1]^2$ represented by the topologist's sine curve, which is connected. But $f$ is clearly not continuous. 
A: Hint: The map $x\to (x,f(x))$ is a continuous map from $X$ to $X\times Y.$
A: One reason your argument is wrong is the line "the disconnectedness of $U$ and $V$ implies $B$ and $D$ are disjoint".
Let $X = ([-1, 1] \setminus 0) \times (-1, 1)$, and consider $$U = [-1, 0) \times (-1, 1), V = (0, 1] \times (-1, 1)$$ as a cover of $X$. These are disjoint and cover the space, but their second components don't fail to be disjoint. (Indeed, they're the same!)
