The proof we give is very elementary and avoids the usual argument seen for this fact, which involves either the Baire Category Theorem or the Hahn-Banach Theorem. It is due to the Chinese mathematician Nam-Kiu Tsing (1984).
Proposition 5.1. No infinite-dimensional normed linear space with a countable Hamel basis can be complete.
Proof. Let $(X,\|\cdot\|)$ be a normed linear space with Hamel basis $(e_n)_n$, and note that without loss of generality we can assume $\|e_n\|=1$.
Let $S_{n-1}$ denote the linear subspace of $X$ spanned by $\{e_1,e_2,\dots,e_{n-1}\}$, and let $r_n \equiv \inf\{ \|x+e_n\|; x\in S_{n-1}\}$ for any $n\ge2$.
Since $\theta\in S_{n-1}$, it follows that $r_n\le \|0+e_n\| =1$ for all $n\ge2$. Now since $S_{n-1}$ is finite-dimensional, it is complete, and hence closed in $X$. Since $e_n\notin S_{n-1}$, we have that $r_n>0$ for all $n\ge2$. Now define the following scalar sequence $(t_n)_n$ by $t_1=1$, $t_2=\frac13$ and for $n\ge2$, $t_{n+1}=\frac13r_nt_n$. Then note that we have
$$0<t_{n+k}\le\left(\frac13\right)^k r_nt_n \le \left(\frac13\right)^{n+k-1}$$
for all $n\ge2$ and $k\in\mathbb N$. Now for each $n\in\mathbb N$, define $u_n=\sum_{i=1}^n t_i e_i$ and note that $(u_n)_n$ is a cauchy sequence in $X$.
But also notice that for any element $u=\sum_{i=1}^{m-1} \alpha_i e_i\in X$, we have
\begin{align}
\| u_{n} - u \|
& = \left\|
\sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m} +
\sum_{i = m + 1}^{n} t_{i} e_{i}
\right\| \\
& \ge \left\|
\sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m}
\right\| -
\left\| \sum_{i = m + 1}^{n} t_{i} e_{i} \right\| \\
& \ge t_{m}
\left\|
\sum_{i = 1}^{m - 1} \frac{1}{t_{m}} (t_{i} - \alpha_{i}) e_{i} + e_{m}
\right\| -
\sum_{i = m + 1}^{n} t_{i}\\
& \ge t_{m} r_{m} -
\sum_{i = 1}^{n - m} \left( \frac{1}{3} \right)^{i} r_{m} t_{m} \\
& \ge \frac{1}{2} t_{m} r_{m}
\end{align}
for all $n>m$. But this means that $\|u_n-u\|$ does not go to zero, which implies $(u_n)_n$ does not converge. Hence, $X$ is not complete.
Tsing N.K. [1984]. Infinite dimensional Banach spaces must have uncountable basis—an elementary proof. Amer. Math. Monthly, 96 (5), 505-506. JSTOR
The symbol $\theta$ is used to denote the zero vector of the space $X$.
Best Answer
If $Y$ has a countable Hamel basis $\{e_1,e_2,\ldots\}$, then let $A=\operatorname{span}\bigl(\{e_1,\ldots,e_n\}\bigr)$. Then $Y=\bigcup_{n\in\Bbb N}A_n$. But each $A_n$ is closed and $\mathring{A_n}=\emptyset$ . So, $Y$ is a countable union of closed subsets with empty interior, and so $Y$ is first category.
And, as you wrote, if it was closed, it would be complete.