Suppose $R$ is a partial order on $A$ and $B \subseteq A$. Prove that $R \cap (B\times B) $ is a partial order on $B$

Question

Suppose $R$ is a partial order on $A$ and $B \subseteq A$. Prove that $R \cap (B\times B) $ is a partial order on $B$.

My attempt:

We need to prove:

1. $R \cap (B \times B)$ is reflexive
2. $R \cap (B \times B)$ is transitive
3. $R \cap (B \times B)$ is antisymmetric.

1.

Take $b \in B$.

Since $B \subseteq A$, and $R$ is partial order on $A$, we have $(b,b) \in R$.

Since $b \in B$, we have $(b,b) \in B \times B$.

Therefore, $(b,b) \in R \cap (B \times B)$.

Since $b$ was arbitrary element, we conclude that $R \cap (B \times B)$ is reflexive.

2.

Suppose we have $x,y,z \in B$ such that $(x,y), (y,z) \in R \cap (B \times B)$.

Since $(x,y), (y,z) \in R \cap (B \times B)$, we know that $(x,y), (y,z) \in R$. And because $R$ is partial order, we can conclude that $(x,z) \in R$

Since $x,z \in B$, we have $(x,z) \in B \times B$

Since $(x,z) \in R$ and $(x,z) \in B \times B$, we have our result

$$(x,z) \in R \cap B \times B$$

Arbitrary elements were considered, thus $R \cap B \times B$ is transitive.

3.

Suppose we have elements $x,y \in B$ such that $(x,y),(y,x) \in R \cap (B \times B)$

We know that $(x,y) \in R$ and $(y,x) \in R$. We know that $R$ is partial order, which means it's antisymmetric. It follows that $x = y$. Arbitrary elements were considered, hence $R \cap (B \times B)$ is antisymmetric.

We've shown that $R \cap (B \times B)$ is reflexive, transitive, and antisymmetric, hence we can conclude that it is a partial order on $B$. $\Box$

Is it correct?