How to prove that a seminorm defines a norm? 
Let $X$ be a vector space, $\|\cdot\|$ a seminorm in $X$ and $M=\left\{x\in X; \|x\|=0\right\}$. Define the relation of equivalance $x\sim y$, if $x-y\in M$. Denote by $[x]$ the equivalence class of $x$ given by $\sim$ and $X/M=\left\{[x];x\in X\right\}.$ Prove that
$$|[x]|:= \|x\|$$
its in fact, a norm in $X/M$.

i) Given $[x],[y]\in X/M$,
$$|[x]+[y]|:=\|x+y\|\leq \|x\| + \|y\|;$$
ii) Given $\lambda \in \mathbb{K}(\mathbb{R}\ or\ \mathbb{C})$
$$|\lambda [x]|:=\|\lambda x\|\leq |\lambda|\cdot \|x\|;$$
iii) $$|x|:=\|x\|=0 \Leftrightarrow x=\vec{0}\in X/M$$
because if $\|x\|=0$ e $x\neq\vec{0}$, then $x\in M$ and that is a contradiction.
I have some difficulties working with equivalence class. Need help.
 A: 
Let $ X $  be a vector space, $\|•\|$  a seminorm in $X$  and
$M=\{x\in X;\|x\|=0\}$. Define the relation of equivalance $x\sim y$ ,
if x−y∈M. Denote by [x] the equivalence class of $x$ given by ∼ and
$X/M=\{[x];x\in X\}$. Prove that $|[x]|:=\|x\|$ its in fact, a norm in
$X/M$.

$1) Positivity$
$2) Absolute \space  homogeneity$
$3) Triangle \space inequality
$
Follows immediately from the property of seminorm $\|•\|$.
Let us verify definiteness property i.e $|[x]|= 0 $ iff $[x]=M$
Let, $[x]=M$
Then, $|[x]|=||x||=0$ $(\text { as }x\in M) $
Now suppose, $|[x]|=0$
Then, $||x||=0$
$\implies x\in M$
Hence, $|[x]|=x+M=M$
A: First, check that the function $ | [x] | $ is well-defined. For that, we have to verify that for any $ y \sim x $, $ \| x \| = \| y \| $, where $ \| \cdot \| $ is the seminorm. This can be established by observing that $ \| y \| \le \| x \| + \| y - x \| = \| x \| $, as $ \| y - x \| = 0 $ for $ y \sim x $. Similarly, $ \| x \| \le \| y \| $, and hence, we must have $ \| x \| = \| y \| $.
Next, note that the operations $ [x] + [y] $ and $ \lambda [x] $ are defined by $ [x] + [y] = [x + y] $ and $ \lambda [x] = [ \lambda x ] $. To verify that $ [x] + [y] = [x + y] $ is well-defined, consider $ x_1 \sim x $ and $ y_1 \sim y $. Then, $ [x] = [x_1] $ and $ [y] = [y_1] $, and $ (x_1 + y_1) - (x + y) = (x_1 - x) + (y_1 - y) \in M $, since $ M $ is a subspace (because for $ u_1, u_2 \in M $ and any scalar $ \lambda $, $ \| u_1 \| = \| u_2 \| = 0 $, and so $ 0 \le \| \lambda u_1 + u_2 \| \le \| \lambda u_1 \| + \| u_2 \| = | \lambda | \| u_1 \| + \| u_2 \| = 0 $, which means $ \| \lambda u_1 + u_2 \| = 0 $, and hence $ (\lambda u_1 + u_2) \in M $). This implies that $ [x_1 + y_1] = [x + y] $, which established that $ [x] + [y] = [x + y] $ is well-defined. Next, since $ (x_1 - x) \in M $ implies $ ( \lambda x_1 - \lambda x) \in M $, we have that $ \lambda [x] = [ \lambda x ] $ is well-defined.
After being sure of the above definitions, the rest are easy. The properties (i) and (ii) do not require further arguments. But property (iii) is written incorrectly in your question (I am presuming that by $ \vec{0} $, you have meant the null vector in $ X $). It would be
$$ | [x] | = 0 \iff [x] = M .$$
Proceeding to prove it, first assume $ | [x] | = 0 $. Then, by definition, $ \| x \| = 0 $, and for all $ y \sim x $, $ \| y \| = \| x \| = 0 $ (follows from arguments in first paragraph), which means $ y \in M $, and hence $ [x] \subseteq M $. Now for any $ y \in M $, since $ x \in M $ and $ M $ is a subspace, we have $ (y - x) \in M $, which implies $ M \subseteq [x] $. Therefore, $ | [x] | = 0 $ implies $ [x] = M $. On the other hand, since $ \vec{0} \in M $ for $ M $ being a subspace, $$ | M | = \left| \left[ \vec{0} \right] \right| = \left\| \vec{0} \right\| = 0 .$$ Hence, we have established
$$ | [x] | = 0 \iff [x] = M .$$
