Boundness of the operator when Hamel basis vectors are eigenvectors
No.
Say $Y$ is a subspace of $X$ and $Y$ is not closed. Say $B_1$ is a Hamel basis for $Y$ and let $B$ be a Hamel basis for $X$ with $B_1\subset B$.
Let $$\lambda_j=\begin{cases}1,&(e_j\in B_1), \\2,&(e_j\notin B_1).\end{cases}$$
Now if $A$ were continuous then $Y=\{x: Ax=x\}$ would be closed.