I want to prove that if $E$ is an infinite, therefore the set of part $\mathcal P(E)$ is infinite. I know that if $E$ is infinite, so $$\forall x\in E, \{x\}\in \mathcal P(E)$$ and so $\mathcal P(E)$ is infinite. But can I do like this:
If $E$ is infinite, then, there is a subset $A\subset E$ which is countable. Let denote $A=\{x_i\}_{i=1}^\infty$ where $x_i\in E$ for all $i$. Consider a subset $B_n\subset A$ which is finite and contain $n$ element of $A$. We know that $\mathcal P(B_n)=2^n$. Moreover, $$\lim_{n\to\infty } B_n=A$$ and $$\lim_{n\to\infty }\mathcal P(B_n)=\mathcal P(A)\subset \mathcal P(E).$$
Therefore $$\lim_{n\to\infty }|\mathcal P(A)|=\lim |\mathcal P(B_n)|=\lim_{n\to\infty }2^n=\infty .$$
Then, we can conclude that $\mathcal P(A)$ is infinite, and so that $\mathcal P(E)$ is infinite too.
Is it correct ?
You used the symbol $\lim_{n \to +\infty}$ in an improper and intuitive way, but not correctly. The problem is that cardinal numbers are not usual numbers, and the fact that they are ordered is a very subtle thing.
You should have argued as follows:
I want to show that for every $n \in \mathbb{N}$ the cardinality of $\mathcal{P}(E)$ is bigger than $n$.
Let $\{ x_i\} \subseteq E$ an infinite countable subset. Then $$n < 2^n = |\mathcal{P}(\{ x_1, \dots, x_n\})| \leq |\mathcal{P}(E)|$$ where the last inequality follows since $\mathcal{P}(\{ x_1, \dots, x_n\} \subseteq\mathcal{P}(E) $.