The statement is false as written, since you could have $b=0$.
A subset $H$ of $K$ is a subfield if and only if $H$ is a subgroup of $K$ under addition, and the nonzero elements of $H$ are a subgroup of the multiplicative group of nonzero elements of $K$.
Thus, $H\subseteq K$ is a subfield of $K$ if and only if:
- $H\neq\emptyset$ and $H\neq\{0\}$.
- If $a,b\in H$, then $a-b\in H$.
- If $a,b\in H$, $a\neq 0$, $b\neq 0$, then $ab^{-1}\in H$.
Proof. The conditions clearly hold if $H$ is a subfield. Conversely, if $H$ satisfies $H\neq\emptyset$ and 2, then $H$ is a subgroup of $K$. Taking $r\in H-\{0\}$ (possible since $H\neq\{0\}$ and $H\neq\emptyset$), setting $a=b=r$ gives $1\in H$, and then condition 3 shows that $H-\{0\}$ is a (multiplicative) subgroup of $K-\{0\}$. Thus, $H$ is closed under addition, products, additive inverses, nonzero multiplicative inverses, and every nonzero element has an inverse. Thus, $H$ is a field, hence a subfield of $K$. $\Box$