P vs NP characterization confusion
The statement is correct for exactly the reason you started with: $\mathsf P$ is a subset of $\mathsf{NP}$ which means that every problem in $\mathsf P$ is also in $\mathsf{NP}$. You can also go through the definitions and find that a deterministic TM running in polynomial time is merely a special kind of nondeterministic TM running in polynomial time and take it from there.
However, while being true, the statement &quot;MST is in $\mathsf{NP}$&quot; holds less value than the statement &quot;MST is in $\mathsf P$&quot; as complexity theory is, at large, concerned with lower bounds. A decent analogy in my opinion would be that the statement $\pi \geq 3$ is stronger than the statement $\pi \geq 0$ eventhough both are true.