Proof of continuity verification. Consider $f(x)=\begin{cases} x, & \text{if $x$ is rational} \\ \frac{1}{x}, & \text{if $x$ is irrational} \end{cases}$
I proved that $f$ is continuous only at $1$ and $-1$ as follows:
Case $1$: $c=0$.
Case $2$: $c$ is rational number that is different than $0,1$ and $-1$. 
Case $3$: $c$ is irrational number.
And in each case I assumed that $f$ is continuous and manged to arrive to a contradiction by constructing a sequence say $\{x_n\}$ that converges to $c$, but  $\{f(x_n) \}$ either diverges or it converges to a number different than $f(c)$.
And for the case of $1,-1$, I proved that f is continuous by the definition of continuity. 
Is my result correct or did I miss something ? And is there a shortcut or faster method than what I did ?
 A: The approach is sound. Without checking the individual steps, it's difficult to verify the proof beyond this point, or to recommend shortcuts.
I don't see to much to shortcut here anyway. You could find most of the points of discontinuity by using the density of the rationals and the irrationals. For any point $x \in \Bbb{R}$, you know that there is a sequence of rationals $a_n \to x$ and a sequence of irrationals $b_n \to x$. Since $x$ is continuous, $f(a_n) = a_n \to x$. Since $\frac{1}{x}$ is continuous (for $x \neq 0$), $f(b_n) \to \frac{1}{x}$.
If $x \neq \frac{1}{x}$ (and $x \neq 0$), then you have two sequences converging to $x$, that map to sequences converging to different limits. This would contradict $f$ being continuous at $x$.
This doesn't establish that $f$ is continuous at $x = \pm 1$, so more individual verification is needed. The $x = 0$ case needs to be dealt with separately too.
So, I wouldn't consider it much of a shortcut! But it might cut down your argument a little.
A: The limit of $f(x)$ by the rationals is $x$. The limit of $f(x)$ by the irrationals is $\dfrac1x$. So for the limit of $f(x)$ to exist, one must fulfill
$$x=\frac1x.$$

In a neighborhood of $x=1$, $|x-1|$ and $|\frac1x-1|=\frac{|x-1|}x$ are virtually equal, so that a $\delta-\epsilon$ argument is easy. Same around $-1$.
A: You method (if you did it correctly) is sound and solid.
An alternative that is perhaps more direct but not shorter or easier would be an $\epsilon$ delta proof.
Rough argument For any $c$ not equal to $\pm 1$  and any $\delta$ you can find rational $x < c$ and irrational $y> c$ so that $x,y \in (c-\delta, c+\delta)$ so that also; either $1<x<c < y$ if $c > 1$; or $0 < x < c < y < 1$ if $0<c < 1$; or $1 < x < 0 < y < 1$ if $c = 0$; or $-1 < x < c <y$ if $-1 < c < 0$; or $x < c < y < 1$ if $c< -1$.
If we compare the values of $f(x)$ to $f(y)$ to $f(c)$... well, if $c =0$ then $|f(y)-f(c)| = |\frac 1y - 0| = \frac 1y > 1$.  Otherwise $f(c)$ will be on "one side" of $\pm 1$ whereas one or the other of $f(x)$ or $f(y)$ will be on the other so that will be a $|f(x|y) - f(c)| > |c -(\pm 1)|$.
So for any  $\epsilon$ less than $|c-(\pm 1)|$ there is no $\delta$ that can guarantee that $|w - c|< \delta \implies 
|f(w) - f(c)|< \epsilon$.
So $f$ is not continuous at $c$.
Of course there are cases to be made. ($c$ rational vs. $c$ irrational and $c= 0$ vs. $0< |c| < 1$ vs. $|c| > 1$ )  But the argument is the same: you can always find an irrational or rational $w$ so that $f(w) >:< \pm 1$ while rational or irrational $c$ is such that $f(c) <:> \pm 1$.
