Prove or provide an example where this is false: There exists $f: (0,1] \rightarrow R $ that is continuous and "onto" The idea is to explain how this is true, or prove an example that disproves.
There exists $f: (0,1] \rightarrow R$ that is continuous and "onto."
My question is: how do I get in the mindset of finding examples that prove or disprove questions like this? 
(I know that in order to prove that this is true, I need to use the Intermediate value theorem.)
 A: A brief comment on the following component:

My question is: how do I get in the mindset of finding examples that prove or disprove questions like this? 

The short answer is, seeing lots of similar problems.
As a longer answer, I can tell you how I might think about this particular question now, though this need not reflect in any way on how you currently conceive of it:
Can I find a continuous, onto function $f:(0,1]\rightarrow \mathbb{R}$?
Well, I need to somehow stretch things, and the interval end with a $1$ does not give me much space with which to maneuver. So let me see how well I can stretch things from the open-end of the interval: What sort of function will send points near (but not equal to) $0$ far away?
I have seen many questions of this nature, such that, by now, the following occurs to me quite naturally: Send $x$ to $1/x$. The function runs into no issues since $0 \not\in (0,1]$.
Okay: What is the image of this function? A bit of thought indicates that it maps $(0,1]$ to $[1, \infty)$.
This has not quite settled matters: The original question requires sending $(0, 1]$ to $\mathbb{R} = (-\infty, \infty)$. So, what is really on my mind was first to stretch the interval using a continuous function so that only one of the two endpoints of the interval is finite (accomplished!) and now my goal is to think about how to find another continuous function that can send $[1, \infty)$ to $\mathbb{R}$. The underlying thinking for this is that I could then compose the functions to get the desired onto component, and I happen to recall that the composition of continuous functions is continuous. 
Finally, I have a pretty large arsenal of functions in my mind; one of them is the hyperbolic cosine from Calculus, and I happen to recall that $\cosh: \mathbb{R} \rightarrow [1, \infty)$. And so taking the inverse hyperbolic cosine will send $[1, \infty)$ to $\mathbb{R}$ as desired.
Therefore, the composition of these functions, $x \mapsto 1/x$ followed by $x \mapsto \cosh^{-1}(x)$ does the trick!

Answer: Define $f: (0, 1] \rightarrow \mathbb{R}$ by $x \mapsto \cosh^{-1}(1/x)$.

A: Try analysing this function's graph $f(x)=\frac 1 x \sin \frac 1 x, x \in (0, 1]$.
A: Update: There was an unforgiveable mistake in the first version.
Fortunately there exist such functions, e.g.,
$$f(x):={1\over x}\sin{1\over x}\qquad(0<x\leq1)\ .$$
On the other hand, if there would be no such functions, there would be no "counterexample" proving that fact. You would have to come up with a genuine idea that transcends all concrete functions that anyone might think of, e.g., some compactness or connectedness argument.
A: There indeed exists such a function. Best example- 
$f(x) = \frac{1}{x}*sin(\frac{1}{x})$
It is indeed continuous, as a multiplication of continuous functions. 
Continuous functions are always locally bounded in a CLOSED interval- [a,b]
Think that once the interval is open, or half open- (a,b) [a,b) or (a,b] you leave room for the function to "go crazy" like this one does. So although the function is always locally bounded, in closed intervals, the local maximum gets infinitely large as we go closer to zero. 
