0/1 Knapsack - Final Answer

After filling the entire table, where's your answer? You want the maximum value using all $n$ items with capacity at most $W$. That's $\max(dp[n][0], dp[n][1], \ldots, dp[n][W])$. because of how the transition works, $dp[n][W]$ already contains this maximum.

Each state considers "capacity $w$" as "up to $w$", not "exactly $w$". So $dp[n][W]$ is your final answer.