Prove that $R[A \cup B] = R[A] \cup R[B]$, where $R$ is a binary relation. Can someone please verify this?

Prove that $R[A \cup B] = R[A] \cup R[B]$, where $R$ is a binary relation.
Here, $R[C] = \{y: \exists x \in C $ such that $(x,y) \in R\}$

Let $z \in R[A \cup B]$.
Then, $\exists x \in A \cup B$ such that $(x,z) \in R$.
But then, $x \in A$ or $x \in B$.
If $x \in A$, then $z \in R[A]$
If $x \in B$, then $z \in R[B]$
In either case, $z \in R[A] \cup R[B]$
So, $R[A \cup B] \subseteq R[A] \cup R[B]$
Now, let $y \in R[A] \cup R[B]$.
Then, $y \in R[A]$ or $y \in R[B]$
If $y \in R[A]$, then $\exists x \in A$ such that $(x,y) \in R$
If $y \in R[B]$, then $\exists x \in B$ such that $(x,y) \in R$
In either case, $x \in A \cup B$.
Since $(x, y) \in R$, we have $y \in R[A \cup B]$
So, $R[A \cup B] \subseteq R[A] \cup R[B]$
Therefore, $R[A \cup B] = R[A] \cup R[B]$.
 A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Your proof looks fine.
Just for fun, here is a proof in a different style, that I personally find easier to read and design: start with the most complex side of the equality, calculate the elements of that set by expanding the definitions, then simplify using the laws of logic, and finally see where this leads you.
So we calculate, for every $\;y\;$,
$$\calc
y \in R[A] \cup R[B]
\calcop{\equiv}{definition of $\;\cup\;$}
y \in R[A] \lor y \in R[B]
\calcop{\equiv}{definition of $\;[\quad]\;$, twice}
\langle \exists x : x \in A : (x,y) \in R \rangle \lor \langle \exists x : x \in B : (x,y) \in R \rangle
\calcop{\equiv}{logic: merge quantifications -- to simplify}
\langle \exists x : x \in A \lor x \in B : (x,y) \in R \rangle
\calcop{\equiv}{definition of $\;\cup\;$ -- start working towards our goal}
\langle \exists x : x \in A \cup B : (x,y) \in R \rangle
\calcop{\equiv}{definition of $\;[\quad]\;$}
y \in R[A \cup B]
\endcalc$$
By set extensionality, this proves the equality.
