Hi to highlight saz' comment you have to realize that every continuous process is predictable. To find non predictable process you have to come up with bigger space $\Omega$ for this take the space of càdlàg processes.
A very classical process that is not predictable is the Poisson process $N$ as it is càg right continuous with jumps of size one, intuition tells you that you won't be able to predict the process at jump time with any sequence of càg processes, in a way, $N$ at jump times is total surprise (unpredictable with knowledge of immediate past).
For more on this you have to work out all the properties of stopping times and their classification, associated representation of predictable sigma algebra (and others) with stopping time, representation of jumps of processes etc... a good reference is Dellacherie Meyer, or more reasonably Protter's book or the book of He, Wang, Yan "Semimartingale theory and Stochastic Calculus" or the blog of George Lowther ( which is regretfully not active anymore)
Best regards
I find the following example easier to understand:
Consider any probability space $(\Omega,\mathcal{A},\mathbb{P})$. Set $\mathcal{F}_t := \mathcal{A}$ for all $t \geq 0$ and let $A \subseteq [0,T]$ be a set which is not Borel-measurable. The process
$$X_t(\omega) := 1_A(t)$$
takes only values in $\{0,1\}$ and for each fixed $t \geq 0$, we have either $X_t = 1$ or $X_t = 0$. This implies in particular that $\omega \mapsto X_t(\omega)$ is measurable (it is a constant!). On the other hand, we have
$$\{(t,\omega) \in [0,T] \times \Omega; X_t(\omega) = 1\} = A \times \Omega \notin \mathcal{B}(\mathbb{R}) \otimes \mathcal{A}$$
and this means that the mapping
$$X: [0,T] \times \Omega \to \mathbb{R}$$
is not measurable, i.e. $(X_t)_t$ is not progressively measurable.
Best Answer
In discrete time:
Consider independent $(Y_n)_{n\in\Bbb{N}}$ with $$P(Y_n = 0) = P(Y_n = 1) = \frac{1}{2}$$
And define $$X_n := \sum_{k=1}^n Y_k$$ being the sum with the filtration given by $$F_n = \sigma(Y_1,\ldots,Y_n)$$
Then $X_n$ is adapted but not predictable
In continuous time:
Consider a poisson point process $(X_t)_{t\ge 0}$ and the related natural filtration given by $$F_t = \sigma(X_s, s\le t)$$
Then $X_t$ is adapted but not predictable.
You can visualize this by the following: $(X_t)_{t\ge 0}$ is a jump process where the time till the next jump is independent from the elapsed time. So if you just have information about the time $s<t$ you won't know if the process will jump at time $t$ hence it's not predictable. But at time $t$ you know if you had jumped or not so it's adapted.