Can someone please help me? I don't know where to start even these questions.
Let $a_n$ be a sequence, and assume that there is a subsequence $b_k=a_{n_k}$ such that $b_k$ tends to $\infty$. Prove that $\sup(a_n)=\infty$.
Let $a_n$ be a sequence, and assume that $\sup a_n=\infty$. Prove that there is a subsequence $b_k=a_{n_k}$ such that $b_k$ tends to $\infty$.
Assume there is a subsequence $a_{n_k}$ such that $\lim_{k\to\infty}a_{n_k}=\infty$. Then for each $M>0$ we may choose $k$ such that $a_{n_k}>M$. This implies that $\sup_n a_n\geqslant M$. Since $M$ was arbitrary, it follows that $\sup_n a_n = \infty$.
Assume now that $\sup_n a_n=\infty$. Choose $n_1$ such that $a_{n_1}>1$. Given $n_k$ such that $a_{n_k}>k$, choose $n_{k+1}>n_k$ such that $a_{n_{k+1}}>k+1$ (we can do this since $\sup_n a_n=\infty$). Then by construction, $\lim_{k\to\infty}a_{n_k}=\infty$.