Is this condition on continuity extraneous or troublesome? I was trying to motivate the use of open sets for defining continuity (as in topology or metric spaces).
I came to formulate the following definition of continuity for a function $f: X \rightarrow Y$.
Let $f(x) = y$, for $x \in X$ and $y \in Y$
Then $f$ is continuous at $x$, if for all set $M \subset Y$, $y \in M$, there exists a set $N \subset X$, $x \in N$ such that $\textbf{f(N) = M}$.
NOTE: The sets $N$ and $M$ are assumed to be connected.
Will this definition of continuity work (as you can see, it doesn't involve open sets)?
Are there any trivial counterexamples?
 A: The usual definition of $f: X \rightarrow Y$ being continuous is that if $M \subseteq Y$ is open in $Y$, then $f^{-1}M$ is open in $X$.  If you want to talk about continuity, you need to somehow be talking about open sets.  You can't just talk about any old subsets.
Another standard definition is that $f$ is continuous at $x$ if and only if for any open set $M$ of $Y$ containing the point $f(x)$, there exists an open set $N$ of $X$ containing the point $x$, such that $f(N) \subseteq M$.  (in the case $X = Y = \mathbb{R}$, this is equivalent to the $\epsilon$-$\delta$ definition of what it means for a function $\mathbb{R} \rightarrow \mathbb{R}$ to be continuous at a point.  do you see why?)

Fact (try to prove this using the definitions): $f$ is continuous if and only if $f$ is continuous at every $x \in X$.

Some comments on your definition:
-Your first statement (1) is true if and only if $f$ is surjective.  It has nothing to do with $f$ being continuous.  To see what I'm getting at, take the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2$ (let's forget about continuity for a second).  This function isn't surjective.  Let $x = 1/2$, so $y = f(x) = 1/4$.  Let $M$ be the set $(-1, 1)$.  Then $y \in M$, but there doesn't exist any set $N$ containing $1/2$ (in fact, there doesn't exist any subset $N \subseteq \mathbb{R}$ whatsoever) with the property that $f(N) = M$, because the negative numbers in $M$ are not mapped to by anything in $\mathbb{R}$ under the function $f$.  
-Suppose we change your first statement (1) to the definition I gave for $f$ being continuous at $x$.  Then your second statement (2) is false.  
A: Here is a counterexample to the following claim: Let $X, Y$ be topological spaces with $x \in X$, and $f: X \rightarrow Y$ a function with the following property: if $M \subseteq Y$ is a connected set containing $y$, then there exists a connected set $N \subseteq X$ containing $x$, such that $M = f(N)$.  Then $f$ is continuous at $x$.
Let $X$ be the unit interval $[0,1]$ (in the subspace topology from $\mathbb{R}$) and let $Y = \{0, 1\}$ with the discrete topology (all subsets are open).  Then the only connected subsets of $Y$ are $\{0\}$ and $\{1\}$.  Define $f:[0,1] \rightarrow Y$ by $f(0) = 0$, and $f(x) = 1$ for $x \in (0,1]$.  Then $f$ is not continuous at $0$, but it satisfies the property above for $x = 0$.  Also $y = 0$, and the only connected set $M$ containing $y$ is $\{0\}$.  And $N = \{0\}$ is a connected subset of $[0,1]$ containing $x$ with the property that $f(N) = M$.  So your definition is satisfied by things which aren't continuous.  
