$(\varepsilon, \delta)$ proof for $f(x)=\begin{cases} \frac{3-x}{2} & x<1 \\ x & x \geqslant1 \end{cases} $ 
Show using $(\varepsilon, \delta)$- definition of continuity that 
  $$f(x)=\begin{cases} 
      \frac{3-x}{2} & x<1 \\
      x &  x \geqslant1 
   \end{cases}
$$
  is continuous at $x=1$.

I had a problem like this on my class and wasn't aware of the $(\varepsilon, \delta)-$definition for continuity and approached the problem a bit differently.
In order for $f(x)$ to be continuous at $x=1$ we would have to have the left and right-handed limits equal to each other.
Since we have:
$\lim_{x\to1^-} \frac{3-x}{2}=1$ (1)
$\lim_{x\to1^+} x= 1$ (2)
we can proceed on proving this using $(\varepsilon, \delta)$ for the limits.
for (1) we can pick $\delta=2\varepsilon$ and since $|\frac{3-x}{2}-1| = |\frac{-x+1}{2}| =|\frac{x-1}{2}| \overset{\mathrm{(x > 1)}}{=}
 \frac{x-1}{2} < \frac{\delta}{2} = \frac{2\varepsilon}{2} =\varepsilon$ the limit holds.
similarly, for (2) we can pick $\delta=\varepsilon$ and since $|x-1|< \delta=\epsilon$ the limit holds as well.
I know this isn't what they asked for, but shouldn't it be pretty much the same thing?

I would say that you are almost done with the problem, but not quite there yet.  What you have is $\delta_\text{left}$ and $\delta_\text{right}$ such that the following holds:  if $0<x-1<\delta_\text{right}$ then $|f(x)-f(1)|<\varepsilon$, and if $0<1-x<\delta_\text{left}$ then $|f(x)-f(1)|<\varepsilon$.  All that's left is finding some $\delta$ such that $|f(x)-f(1)|<\varepsilon$ whenever $|x-1|<\delta$.
