If the diameter of graph is greater than 3 then the diameter of its complement graph is less than 3

Denote distance in $G$ by $d$ and distance in $G'$ by $d'$. Define $u,v$ as you did, so that $d(u,v) \ge 4$. We wish to show for any vertices $x,y$, that $d'(x,y) \le 2$.

Among $x, y, u, v$ there are at most four distinct vertices. In particular, since $d(u,v) \ge 4$, $u$ and $v$ cannot be connected by a path among $x,y,u,v$. So it is possible to split $x,y,u,v$ into two components which are not connected in $G$, one containing $u$ and one containing $v$. WLOG we may assume the components are $\{u,x\}, \{v,y\}$ or $\{u,x,y\}, \{v\}$.

• In the first case, $x$ and $y$ have no edge in $G$, so they have an edge in $G'$ and $d'(x,y) = 1$.

• In the second case, $x$ and $y$ are both connected to $v$ in $G'$, so $d'(x,y) \le d'(x,v) + d'(y,v) = 2$.

Thus $d'(x,y) \le 2$ for any $x,y$, proving that $\operatorname{diam} G' \le 2$.

Denote by $d_G$ and $d_{G'}$ the distances in $G$ and $G'$, respectively. Take vertices $u,v$. There are a few cases of interest.

If $d_G(u,v)\geq 2$, then the edge $uv$ is not in $G$, so $uv\in G'$ and thus $d_{G'}(u,v)=1$.

Now suppose $d_G(u,v)=1$. Take other vertices $p,q$ with $d_G(p,q)\geq 3$ (this must exist as diameter $\geq 3$), so in particular one of $p$ or $q$ is not connected to $u$. Similarly, one of $p$ and $q$ is not connected to $v$. There are two cases:

Both $u$ and $v$ are not connected (in $G$) to the same vertex, say $p$. Then $upv$ is a path of length $2$ in $G'$ from $u$ to $v$ and thus $d_{G'}(u,v)=2$.

$u$ is not connected to $p$ and $v$ is not connected to $q$ (or vice-versa). Since $p$ is also not connected to $q$, we have a path $upqv$ in $G'$ from $u$ to $v$ and thus $d_{G'}(u,v)\leq 3$.

Therefore, $G'$ has diameter at most $3$.