real analysis – Prove that $limsup a_nleq sup a_n$

So here’s one question which I came across some time ago.

Given any sequence $a_n$ in $mathbb{R}$, I need to show that $limsup a_nleqsup a_n$.

My approach: As the sequence $a_ninmathbb{R}$, thus to talk about the $limsup a_n$, I assumed that $a_n$ is bounded. I defined
$$b_n:=sup{a_k:kgeq n}$$ i.e.
$$begin{align}b_1&=sup{a_1,a_2,…}\b_2&=sup{a_2,a_3,…}\b_3&=sup{a_3,a_4,…}\.&\.end{align}$$
Now, I can say that $<b_n>$ is a decreasing sequence. Thus, we have, $b_1geq b_2geq b_3geq …$ and so on.

Now, like we had defined $b_n$, we get that, $sup a_n=b_1$.

I then did this:
$$sup b_1=sup{b_1,b_2,…}geqinf{b_1,b_2,…}impliessup a_ngeqlimsup a_n$$
The reason why I did this is because $<b_n>$ is a decreasing sequence. But I am a bit doubtful if I can write $sup b_1=sup{b_1,b_2,…}$. Can anyone please verify this?

Also, is there any alternate way of proving this inequality? Any help would be very much appreciable.