Q:

proof that f*d is an induced metric

Let X be a set and (M, d) be a metric space. Let f : X → M be an injective map. Then we define
f∗d := d ◦ (f × f): X × X →(o,infinity] 
i.e.
f∗d(x, y) = d(f(x), f(y)) for x, y ∈ X .
I want to prove that f∗d is a distance function on X (“induced metric”). here's what ive tried so far ...
taking all of the above as fact ;
f is injective therefore f(x)=m1 and f(y)=m2  where m1 does not equal m2 and m1,m2 are elements of M. 
=> fd(x,y)=d(m1,m2) which is defined already as a metric space X C(M,d) therefore fd is also a distance function.
...I feel like thats too easy and i must be missing something obvious ? any tips would be much appreciated but please dont answer the full question as i want to learn this on my own :)  

A:

$f^*d(x,y)=d(m_1,m_2)$ which is defined already as a metric space ... therefore $f^*d$ is also a distance function.

This sentence doesn't make any sense.  To what is the “which” referring to?  The only possible antecedent in the sentence is the number $f^*d(x,y)=d(m_1,m_2)$.  A number isn't defined as a metric space.
If you're asserting that the facts that $d$ is a metric and $f^*d$ is defined in terms of $d$, make $f^*d$ a metric on $X$—well, that's exactly what you need to prove.
Go back to the definitions of metric:

You must show $f^*d(x,y) \geq 0$ for all $x,y \in X$, and if $f^*d(x,y) = 0$, then $x=y$.
You must show $f^*d(x,y) = f^*d(y,x)$ for all $x,y \in X$.
You must show the triangle inequality: For all $x$, $y$, $z\in X$, 
$$
    f^*d(x,y) + f^*d(y,z) \geq f^*d(x,z)
$$

Each of those statements can be expressed in terms of $f$ and $d$. You'll know you're doing it right when you find that you need to use each of the requirements that must be satisfied for $d$ to be a metric, and the assumption that $f$ is injective.....

