If $f:M \mapsto M'$ is bijective and $d'(f(x),f(y))\ge d(x,y)~\forall x,y\in M$ then $M$ compact $\implies M'$ compact Let $(M,d)$ and $(M',d')$ be metric spaces and let $f:M\mapsto M'$ be a bijective function such that $$d'(f(x),f(y))\ge d(x,y)~\forall x,y\in M$$
Is it true that if $M$ is compact then so is $M'$?
I found that the function $f(x)=\begin{cases} x & \text{ if } x\in[0,1/2)\\ x+1/2 & \text{ if } x\in[1/2,1]\end{cases}$ satisfies the conditions for $M=[0,1]$ and $M'=[0,1/2)\cup[1,3/2]$ and $M'$ is not compact. 
Is this true? What if $f$ had to be continuous?

Yes, your example works. If you're worried, then try going into more detail. How do you know $M$ is compact? (Cite a theorem.) How do you know $M'$ is not compact? (Find a sequence which contains no convergent subsequence, or indeed a non-convergent Cauchy sequence would do!) Are you sure that $d(x, y) \le d'(f(x), f(y))$? Prove it!
If $f$ is continuous, then $M'$ must be compact, as the continuous image of a compact set is compact (regardless of whether $d(x, y) \le d'(f(x), f(y))$ holds true or not).
