Solution
= Solution
\b[True.] The same subsequence argument applies to the <sum of a compact set and a closed set>. For a convergent sequence $z_j=x_j+y_j$, boundedness of $X$ provides a convergent subsequence of $x_j$, with limit $x\in X$. Then $y_j\to z-x\in Y$ along that subsequence. Consequently $z\in X+Y$. No boundedness of $Y$ is needed.