Something that isn't continuous can be proven to be continuous (so it is continuous - definitions - but doesn't look it!) I'm sorry to post this, either I am right and it is continuous, or because I am on $\mathbb{Q}$ not $\mathbb{R}$ that saying "if that delta works, any smaller delta will!" (which can be proven by by some * value theorem) does not work.
Consider this, now I was trying to prove it isn't continuous, I am now convinced it is. $f:\mathbb{Q}\rightarrow\{1,2\}$ (I chose 1 and 2 to make the sketch nicer) given by $f(x)=1$ if $x<\sqrt{2}$ and $f(x)=2$ otherwise/$x>\sqrt{2}$ - it is important that this change happen about a number in $\mathbb{R}$ but not $\mathbb{Q}$.
Now I remember when proving the continuity of 1/x over a year ago, I learnt that it is helpful (necessary) to bound delta above when you have points of discontinuity, this stops it getting too close, it also keeps it to one side of the discontinuous point.
I seek to prove $\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon$ to mean continuous at $a$.
Now that upper bound I mentioned, because this function is flat I don't need to find a smaller delta that's a function of epsilon and take the minimum. The proof is trivial.
Let $\delta=|\sqrt{2}-a|$, now if $|x-a|<\delta$ I am saying in words "The distance from x to a is less than the distance from a to that nasty point" which means $x\in(a-\delta,a+\delta)$ which is clearly a ... chunk of the domain either entirely before that $\sqrt{2}$ or after.
On this $f(x)-f(a)=0$ which is less than $\epsilon$ for all $\epsilon>0$, thus I have proved that this function is continuous. EVEN though it has a jump in it! It is continuous on $\mathbb{Q}$
What I think I have learnt
I think I have learnt that while 1 and 2 seem far apart for real numbers, or even fractions, in the set {1,2} there is no middle value. So the jump is not actually a jump at all. 
With this I am not sure if the $\sqrt{2}$ thing is actually significant. If we had a function that was 1 if $x\le\frac{1}{2}$ say, else 2. does this have the "no-jump" quality? I can see a case for no, if $\epsilon<1$  then no it cannot be continuous, because there can be a change near $x=\frac{1}{2}$ by a value of more than one, no matter how small $\delta$ (at x=0.5).
HOWEVER it might be yes. If you say "the change must be -1,0 or 1" thus confining $\epsilon$ to take 1, it is still no (as 1 is not less than 1) but one could question whether it is fair to try and impose this "less than" on {1,2} in this way. If it ever changes there can be no smaller jump. I am reading about the issue, I'm posing this because my foundations have somewhat crumbled, I'd like some help patching them back up.
My apologies for the naff format/style of this question, it suffers from me thinking about what I am writing and flicking between several different ways and making sure it is consistent. I've read it twice and it is awful, but I cannot think how else to phrase it. 
 A: Your proof is correct.  The function $f$ is continuous, and your argument that "if this $\delta$ works then so will any smaller $\delta$" is perfectly sound.
This example does tend to challenge people's intuition about continuity.  One way I like to think about it is that continuity is a pointwise property.  If you want to claim that a function $f : X \to Y$ is not continuous, you have to exhibit a point $x \in X$ where that discontinuity occurs.  Here with $X = \mathbb{Q}$, you would like to say the discontinuity occurs at $x = \sqrt{2}$, but of course that isn't a point of $\mathbb{Q}$.  Indeed, there's no point of $\mathbb{Q}$ where a discontinuity occurs, so the function must be continuous.
One way of addressing this issue is via the concept of uniform continuity.  It would be a good exercise to show that, although your function $f$ is continuous, it is not uniformly continuous.  Indeed, another great exercise is to show that a function $f : \mathbb{Q} \to \mathbb{R}$ is uniformly continuous if and only if it has a continuous extension to $\mathbb{R}$, i.e., $f = g|_{\mathbb{Q}}$ for some continuous $g : \mathbb{R} \to \mathbb{R}$.  In some sense, uniform continuity is able to detect jumps that occur at "holes" in the domain.
A: A more simple-minded example to illustrate the point is the function $f(x)=\frac{1}{x}$.  Note that $f$ is continuous at every point where it is defined. If you insist on assigning a value to $f$ at the origin, the resulting function will necessarily be discontinuous. Similarly the function you mentioned is continuous everywhere it is defined. If you assign a value to it at $\sqrt{2}$ it will become discontinuous.
Edit 1. I just noticed that @JiK made the same point in a comment. If you wish you can format your comment as an answer and I will delete mine.
