Prove that, if a code $C\subseteq \left \{ 0,1 \right \}^n$ has code distance (minimum Hamming distance between pairs of codewords) $d$, then the code $C'=\left \{ (u,u):u\in C \right \}$ has code distance $2d$.
The code is a subset of the set of all possible binary strings of length $n$, i.e. $\left \{ 0,1 \right \}^n$. The elements of this subset are called codewords.
My attempt: We need to show that the minimum distance between any two different code elements $C'$ is $2d$. Suppose that there exist two different elements $(u_1,u_1)$ and $(u_2,u_2)$ of $C'$ such that the distance between them is less than $2d$. Then the distance between $u_1$ and $u_2$ must be less than $d$, because any code word in $C'$ is of the form $(u,u)$, where $u$ is the code word in $C$. This contradicts the fact that $C$ has a codeword distance of $d$. So any two different elements of $C'$ must have a distance of at least $2d$.
Is my reasoning and the proof itself correct?
Your proof is almost correct; now you need to show the minimum distance is no more than $2d$. This doesn't take too much more work, though :)
You know that the minimum distance the smallest pairwise distance, so if you take any two elements and compute the distance between them, the minimum distance of the code $C$ is smaller (by definition). Well, you know that there exists some pair of codewords $c_1, c_2 \in C$ which are only distance $d$ from each other, since the minimum distance of $C$ is $d$.
Since that's the case, now consider the codewords $(c_1,c_1),(c_2,c_2) \in C'$. They are exactly distance $2d$ from each other. Since you have found two codewords in $c'$ which are distance $2d$ from each other, you now know the minimum distance of $C'$ must be at most $2d$.
This, combined with the work you have in your solution, shows the minimum distance of $C'$ is exactly $2d$.