There are no two words of a code $C$ that have weight $\leq \lfloor \frac{d-1}{2} \rfloor$.
Proof:
$e_1, e_2 \in x+C$ with $||e_1||, ||e_2|| \leq \lfloor \frac{d-1}{2} \rfloor$
then $||e_1-e_2|| \leq ||e_1||+||e_2||\leq 2 \lfloor \frac{d-1}{2}\rfloor \leq d-1$, contradiction.
Could you explain to me why this is a contradiction? The minimum weight of an element is $\lfloor \frac{d-1}{2}\rfloor$ and $d-1 \geq \lfloor \frac{d-1}{2}\rfloor$. So why can't we have a codeword of weight $d-1$ ?
EDIT: If we calculate $x+C$ we will not get the same words as these of the code, but we will know that the word with weight $\leq \lfloor \frac{d-1}{2}\rfloor$ is the error since $C$ cannot have words with weight $\leq \lfloor \frac{d-1}{2}\rfloor$ and so $x+C$ should also not have? Or isn't it like that?