Let the supremum subsequential limit be $L$. Assume the contrary, that $L< \lim \sup$. Write $L +2 \epsilon= \lim \sup$ for $\epsilon >0$.
If there are only finitely many $i$ such that $a_i \geq L+ \epsilon$, then consider the maximum such $i$, say $M$. Then $\forall n\geq M+1$, $a_n <L+\epsilon \implies \forall u >M, \ \sup_{n \geq u} a_n \leq L+ \epsilon \implies \lim \sup \leq L +\epsilon$. Contradiction.
If there are infinitely many $i$ such that $a_i \geq L+ \epsilon$, then construct a sequence of such $a_i$s say $(a_{k_i})$. Of course, $a_i$ is bounded above so $(a_{k_i})$ is a bounded sequence. By Bolzano Weierstrass, it has a convergent subsequence, whose limit $\ell \geq L+ \epsilon > L$. Contradiction since $L$ is the supremum subsequential limit.
We will also show the supremum subsequential limit exists. Take any convergent subsequence $(a_{k_i}) \rightarrow t$. Since $k_i \geq i$ for all $i$, $\sup_{n \geq i}a_n \geq a_{n_i}\ \ \forall i$. This implies $\lim \sup \geq t$. So all subsequential limits are bounded above by $\lim \sup \implies$ $L$ exists and $\lim \sup \geq L$. Combining with the previous inequality, $L = \lim \sup$.