Is the proof $diam(A\cup B) \le diam(A) +diam(B)+d(A, B)$ logically perfect

$(X, d) $ be a metric space and $A, B$ be two non empty bounded subsets of $X$.

To show : $$diam(A\cup B) \le diam(A) +diam(B)+d(A, B)$$

Notations:

$diam(A) =Sup\{d(x,y):x,y\in A\}$

$d(A,B) = Inf\{d(x,y): x\in A, y\in B\}$

My attempt:
$(X, d) $ metric space.$ x, y\in X$

$diam(A\cup B) =sup\{d(x, y) : x, y \in A\cup B\}$

$x, y\in A\cup B $ this implies both $x, y \in A $ or $x, y \in B$ or, one of them belongs to $A$ and other belongs to $B$.

We will consider two cases

Case 1: Both $x, y\in A $ ,(similarly both $x,y \in B$) then

\begin{align} d(x, y)&\le\sup\{d(x, y):x, y \in A\}\\
&=diam(A) \end{align}

$\begin{align} diam(A)&\le diam(A) +diam(B) +d(A, B)\end{align}$

Hence,\begin{align} diam(A) +diam(B) +d(A, B)\end{align}
is an upper bound of $\{d(x,y):x,y\in A\cup B\}.$

And hence, \begin{align} diam (A\cup B) &=sup\{d(x,y):x,y\in A\cup B\}\\
&\le diam(A) +diam(B) +d(A, B) \end{align}

And similarly, $x, y\in B \implies diam(A\cup B \le diam(B)\le diam(A) +diam(B) +d(A, B) $

Case 2: $x, y$ doesn't belongs to the same set.(A Or B)

W.L.O.G we asumme
$x\in A \text{ , }y\in B $, then choose two points $a\in A, b \in B. $

\begin{align}
d(x, y) &\le d(x, a) +d(a, b)+d(b, y)\\
&\le diam(A) +diam(B) +d(a, b) \end{align}

Hence, $d(x, y) – diam(A) – diam(B) $ is a lower bound of$ \{d(a,b):a\in A,b\in B \}$.

So, \begin{align}d(x, y)-diam(A) -diam(B) &\le Inf \{d(a,b):a\in
A,b\in B \} \\
&\le d(A, B)
\end{align}

Hence, $ diam(A) +diam(B) +d(A, B)$ is an upper bound of the set $\{d(x,y):x,y\in A\cup B\}$

And so,\begin{align}diam(A\cup B) &= sup\{d(x,y):x,y\in A\cup B\}\\
&\le diam(A) +diam(B) +d(A, B)\end{align}.

I want to know , is there any logical gaps?

Is my proof correct?

How many marks do you want to give me out of 5 ?

Please verify the proof. Thanks.