Is every bounded sequence convergent?
No. For example, take the sequence $$a_n=\begin{cases} 0 & \text{ if }n\text{ is even}\\ 1 & \text{ if }n\text{ is odd} \end{cases}$$ It is bounded because it stays inside the interval $[0,1]$, but it has no limit.
Intuitively, you shouldn't expect that bounded $\implies$ convergent, because even if the terms of a sequence stay in some general area, doesn't mean that all of its terms must always be getting closer and closer to each other (which is what the notion of Cauchy sequence captures; a sequence in $\mathbb{R}$ or $\mathbb{C}$ is convergent $\iff$ it is Cauchy).
However, as user amWhy points out in their answer, every bounded sequence contains a convergent subsequence; in other words, we can pick out some terms of the sequence that are getting closer and closer to each other (even if they aren't getting closer to all the terms in the original sequence).
Hint: $\quad$Consider the sequence $\{a_n\},\;a_n = (-1)^n\,$
It is bounded in $[-1, 1]\; ($ indeed, $a_n \in \{-1, 1\}\; \forall a_n\in \{a_n\}),\;$ but $\;\lim_{n\to \infty} (-1)^n\;$ does not exist.
Note: it is true that every bounded sequence contains a convergent subsequence, and furthermore, every monotonic sequence converges if and only if it is bounded.
Added See the entry on the Monotone Convergence Theorem for more information on the guaranteed convergence of bounded monotone sequences.
No. Take the sequence $a_n=i^n$ in the complex plane. It is bounded since it is contained in the circle $|z|\leq \sqrt{2}$. However it doesnt converge since its terms alternate depending on $n$ modulo 4.