Our professor showed us Bolzano-Weierstrass theorem for sets only (every infinite, bounded set $A\in\mathbb{R}$ has at least one accumulation point in $\mathbb{R}$) and left us to prove that every sequence of real numbers has at least one limit point and I would be grateful if anyone would tell me if my proof is correct:
Sequence is infinite and bounded.
Let $A=\{x_n|n \in\mathbb{N}\}.$ Since $A$ is both bounded and infinite existence of limit point comes directly from BW theorem for sets
Sequence is infinite and unbounded.
Let $G$ be some neighbourhood of $+\infty$ (same applies for $-\infty$). For any $M\in\mathbb{R}, \exists n\in\mathbb{N}$ such that $x_n\in(M,+\infty)$ $\forall n\geq$ some $n_0$ thus there is a subsequence of $x_n$ that converges to infinity and so we can say that $+\infty$ is limit point of $x_n$
Sequence is finite and bounded
There is certain real $a$ such that $x_n=a$ for finite $n$.$\implies \exists x_{n_k}=a; \forall k\in\mathbb{N}\implies lim_{k\to\infty} x_{n_k} = a$ thus there is subsequence of $x_n$ that converges to some point ($a$) which is its limit point.
Sequence cannot be finite and unbounded in $\mathbb{R}$
I was looking also through lots of previous questions on this site that were somewhat similar but none of them have quite answered my question or I didn't understand the solution.
Please correct me if I am wrong somewhere I spent couple of hours to first find connections between BWT for sets and sequences and then prove this... Limits superior and inferior are introduced in later lectures so I am not allowed to use them.