From the notes I'm studying from I read that " $\mathbb{Z}\_p=\mathbb{F}\_p$ has no proper subfield."

The rationale is: "assuming $\mathbb{K}$ is a subfield of a finite field $\mathbb{Z}\_p= \mathbb{F}\_p$, $p$ prime, then $\mathbb{K}$
must contain 0, and 1, and so all other elements of $\mathbb{F}\_p$ by the closure of $\mathbb{K}$ under
addition. Therefore, it follows that $\mathbb{Z}\_p=\mathbb{F}\_p$ contains no proper subfield"

But thinking about $\mathbb{Z}\_5 = \{0,1,2,3,4\}$ and $\mathbb{Z}\_3 = \{0,1,2\}$

Isn't $\mathbb{Z}\_3$ a proper subfield of $\mathbb{Z}\_5$ since it contains 0, 1 and is closed under addition and multiplication?