Let be $ A $ a set of convergent sequences. Suppose $$ b_n:=\sup\{a_n:(a_n)_{n\in \mathbb{N}}\in A\} $$ does exist for all $ n\in \mathbb{N} $. Then for $ (b_n)_{n\in \mathbb{N}} $ follws $$ \lim\limits_{n\to \infty} b_n=\sup\{\lim\limits_{n\to\infty} a_n:(a_n)_{n\in \mathbb{N}}\in A\} $$.
My idea:
Let be $ A $ a set of convergent sequences described by $$ A:=\{(a_n)_{n\in \mathbb{N}}:(a_n)_{n\in \mathbb{N}} \text{ converges}\}. $$ So for all $ (a_n)_{n\in \mathbb{N}}\in A $ applies the following $$ \forall \varepsilon>0 \ \ \exists N\in \mathbb{N} \ \ \forall n\geq N: |a_n-a|<\varepsilon, $$ and $ a-\varepsilon<a_n<a+\varepsilon $ with $ a:=\lim\limits_{n\to \infty} a_n $. Suppose $$ b_n:=\sup\{a_n:(a_n)_{n\in \mathbb{N}}\in A\} $$ does exist for all $ n\in \mathbb{N} $. Because of $$ \left(\lim\limits_{n\to \infty} a_n\right)-\varepsilon=a-\varepsilon<a_n $$
the value $$ \sup\{a:(a_n)_{n\in \mathbb{N}}\in A\} $$ also does exist. Then $$ \begin{align}&\sup\{a:(a_n)_{n\in \mathbb{N}}\in A\}-\varepsilon\\&=\sup\{a-\varepsilon:(a_n)_{n\in \mathbb{N}}\in A\}\\&<\sup\{a_n:(a_n)_{n\in \mathbb{N}}\in A\}\\&<\sup\{a+\varepsilon:(a_n)_{n\in \mathbb{N}}\in A\}\\&=\sup\{a:(a_n)_{n\in \mathbb{N}}\in A\}+\varepsilon\end{align} $$ By Squeeze theorem we get $$ \lim\limits_{n\to \infty} b_n=\sup\{\lim\limits_{n\to\infty} a_n:(a_n)_{n\in \mathbb{N}}\in A\}. $$
Does this work?