I found a proof " The $\mathbb{Z}$ - module $M=\mathbb{Z} \times\mathbb{Z} \times\dots$ is not projective" in T.Y.Lam - Exercises in Modules and Rings.
Proof
Suppose that $M$ is projective, so $M$ is isomorphic to a direct summand of a free module then $M \subseteq F$ for a suitable free abelian group $F$ with basis $\{e_i| i \in I\}$. We have $P=\mathbb{Z}\oplus \mathbb{Z}\oplus \dots\oplus \mathbb{Z}\subseteq M$ be countable, so we can decompose $I$ into a disjoint union $I_1\cup I_2$ such that $I_1$ is countable and $P$ is contained in the span $F_1$ of $\{e_i|i\in I_1\}$. Let $F_2$ is the span of $\{e_i|i\in I_2\}$.\ The group $M/(M\cap F_1)$ has an embedding into the free abelian group $F/F_1\cong F_2$. We will get a contradiction if we can show that $M/(M\cap F_1)$ contains a nonzero element $\alpha$ that is divisible by $2^n$ for any $n\ge 1$. Consider the set $$S=\{(2^1\epsilon_1,2^2\epsilon_2,2^3\epsilon_3,\dots):\epsilon_i=\pm1\}\subseteq M$$ We see that $F_1$ is countable but $S$ is uncountable, so $S$ contains $$a=(2^1\epsilon_1,2^2\epsilon_2,2^3\epsilon_3,\dots) \not \in F_1$$ Hence, we have $$\begin{array}{*{20}{l}} \alpha &=& (2{\epsilon_1},{2^2}{\epsilon_2}, \ldots ,{2^{n - 1}}{\epsilon_{n - 1}},0,0, \ldots ) + (0,0, \ldots ,{2^n}{\epsilon_n},{2^{n + 1}}{\epsilon_{n + 1}}, \ldots ) + M \cap {F_1}\\ &=& {2^n}\left( {0,0, \ldots ,{\epsilon_n},2{\epsilon_{n + 1}}, \ldots } \right) + M \cap {F_1} \end{array}$$ Then $\alpha=a+(M\cap F_1)\in M/(M\cap F_1)\setminus \{0\}$ is divisible by $2^n$ for any $n$.
I have a question: Why we get a contradiction when we show that $M/(M\cap F_1)$ contains a nonzero element $\alpha$ that is divisible by $2^n$ for any $n\ge 1$. Thankyou.
You say $M/(M\cap F_1)$ has an embedding into some free Abelian group $F_2$. Let $\alpha\in F_2$. Let $(e_i)_{i\in I}$ be a basis of $F_2$. Then $\alpha =\sum_{i\in I_0} a_i e_i$ where $a_i\in\Bbb Z$ and $I_0$ is a finite subset of $I$. If $\alpha\ne0$ then some $a_i\ne 0$, and in $F_2$, the only positive integers that $\alpha$ could possibly be divided by are the factors of $a_i$, of which there only finitely many. So in $F_2$ no nonzero element is divisible by $2^n$ for all $n$.