Real valued function, $M$, $s_n \rightarrow s; |f(s_n)| \le M$ for $s$ in open $U$. (Problem from Gamelin and Greene) Gamelin and Green, Introduction To Topology Ch. 1 section 3 problem 10.

Let $f$ be a real-valued function  on $\mathbb{R}$, the real numbers.  Show that there exist $M \gt 0$ and a nonempty open subset $U$ of $\mathbb{R} $ such that for any $s \in U$ , there is a sequence $\{s_n\}$ satisfying $s_n \rightarrow s$ and $|f(s_n)| \le M, n \ge 1$.

I intuitively get this.  Every point in an open set has a sequence converging to it, so this is just saying if any $s_n$ are "too big" I can find another $s'_n$ close by that isn't "too big". ($s$, itself, can be as big as it wants.)
This is obviously true if $f$ is continuous or even just bounded on any open interval. (Just let U be the image $f(A)$ of any open interval, $A$, and let $M = \sup f(A)$). For this not to be true, $f$ needs to be unbounded on every possible open set. Not merely unbounded which is easily possible but.... If $V_M =\{x : f(x) \ge M\}$ will always always be open no matter how large the $M$.  Which seems intuitively impossible.  But, I don't have it.
At least I don't with concepts of the first 15 pages of the book. (Metric spaces, closed and open sets, convergent sequences, reals having least upper bound property, but not compactness; it's a pretty dense book.)
 A: New approach!
$x$ is called good for $M$ if there's a sequence $\{x_i\}$ tending to $x$ with all its $f$-values $\le M$.
Clearly every point $x$ is good for some $M$, for example for $f(x)$.
It follows that if we define $G_M = \{x \in R: x \text{ good for } M\}$, then the countable union $G_1\cup G_2 \cup G_3 \cdots$ is all of $\mathbb{R}$.
If only some $G_M$ were open, that would have settled the problem. But actually life isn't that easy. Actually each $G_M$ is closed. Let's prove this. Take $\{x_i\}$ a sequence of points in some $G_M$ tending to $x$. Use the good sequences tending to every $x_i$ and combine them to build a good sequence tending to $x$, therefore $x \in G_M$ (details omitted).
OK, so if $G_M$ is closed, then let's take its interior. If it's nonempty, that will be the open set we need.
But maybe all the closed sets $G_M$, for every $M$, have empty interiors. We will prove that can't be the case using the Baire category theorem.
If all $G_i$, $i=1,2,3,\dots$ have empty interiors, then their complements $U_i = R\setminus G_i$ are dense open sets. Since the union of $G_i$ is $\mathbb{R}$, the intersection of $U_i$'s is empty. But by the Baire category theorem this intersection is a dense (so certainly nonempty) set. Contradiction.
Therefore at least one $G_i$ will have a nonempty interior, and its interior will be the open set we need.
A: EDIT: the approach below doesn't work.
Since you know absolutely nothing about continuity or any kind of nice behavior of $f$, it seems fruitless to try and correlate its behavior with open sets and other such topological objects (what does $f$ care between an open set and some other random continuum-sized subset of $R$?).
But, here's a nice thing, you're only asked to evaluate $f$ on the $s_n$ that you provide - not on $s$. This is important! If we always choose $s_n$ from a very limited stock of points $S$, we only need to care about values of $f$ on those points and nothing else. But as long as $S$ is continuum-sized, $f$ can be just as horrible on $S$ as it is on the whole of $R$.
Hmm, do we have a less-than-continuum-sized subset $S$ of $R$ such that every real has a sequence in $S$ convergent to it? Right!
(I advise to stop reading here and try continue on your own for a while, come back if it doesn't work out).
OK, now take any $x \in R$ and $M > 0$. Maybe $x$ has a sequence of rationals converging to it, in which we can find a subsequence which has all its $f$-values below $M$. In this case let's say that $x$ is a good point for $M$.
Is it possible that there's no pair $(x,M)$ such that $x$ is good for $M$? No; if $x$ itself is rational, it is trivially good for $f(x)$. 
OK, so for some $M$ there are good points. Maybe all the good points for $M$ together make an open set? Hint: look at the bad points for $M$ and try to show it's a closed set.
