Which of the following describes the largest subset of values of $y$ within the closed interval $[0,\pi]$ for which 
$$\sin(x+y)\leq \sin(x)+\sin(y)$$
for every $x$ between $0$ and $\pi$, inclusive?
$\textbf{(A) } y=0 \qquad \textbf{(B) } 0\leq y\leq \frac{\pi}{4} \qquad \textbf{(C) } 0\leq y\leq \frac{\pi}{2} \qquad \textbf{(D) } 0\leq y\leq \frac{3\pi}{4} \qquad \textbf{(E) } 0\leq y\leq \pi $
1. We start with the given inequality:
   \[
   \sin(x+y) \leq \sin(x) + \sin(y)
   \]
   for every \( x \) in the interval \([0, \pi]\).

2. Using the angle addition formula for sine, we can rewrite \(\sin(x+y)\) as:
   \[
   \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)
   \]

3. To compare \(\sin(x+y)\) with \(\sin(x) + \sin(y)\), we need to analyze the expression:
   \[
   \sin(x)\cos(y) + \cos(x)\sin(y) \leq \sin(x) + \sin(y)
   \]

4. We know that \(\cos(x) \leq 1\) and \(\cos(y) \leq 1\) for all \( x \) and \( y \) in \([0, \pi]\). Therefore:
   \[
   \sin(x)\cos(y) \leq \sin(x) \quad \text{and} \quad \cos(x)\sin(y) \leq \sin(y)
   \]

5. Adding these inequalities, we get:
   \[
   \sin(x)\cos(y) + \cos(x)\sin(y) \leq \sin(x) + \sin(y)
   \]

6. This shows that the inequality \(\sin(x+y) \leq \sin(x) + \sin(y)\) holds for all \( x \) in \([0, \pi]\) and \( y \) in \([0, \pi]\).

7. Therefore, the largest subset of values of \( y \) within the closed interval \([0, \pi]\) for which the inequality holds is the entire interval \([0, \pi]\).

Conclusion:
\[
\boxed{0 \leq y \leq \pi}
\]