Example of a stochastic process which does not have the Markov property
An urn contains two red balls and one green ball. One ball was drawn yesterday, one ball was drawn today, and the final ball will be drawn tomorrow. All of the draws are "without replacement".
Suppose you know that today's ball was red, but you have no information about yesterday's ball. The chance that tomorrow's ball will be red is 1/2. That's because the only two remaining outcomes for this random experiment are "r,r,g" and "g,r,r".
On the other hand, if you know that both today and yesterday's balls were red, then you are guaranteed to get a green ball tomorrow.
This discrepancy shows that the probability distribution for tomorrow's color depends not only on the present value, but is also affected by information about the past. This stochastic process of observed colors doesn't have the Markov property.
Update: For any random experiment, there can be several related processes some of which have the Markov property and others that don't.
For instance, if you change sampling "without replacement" to sampling "with replacement" in the urn experiment above, the process of observed colors will have the Markov property.
Another example: if $(X_n)$ is any stochastic process you get a related Markov process by considering the historical process defined by $$H_n=(X_0,X_1,\dots ,X_n).$$ In this setup, the Markov property is trivially fulfilled since the current state includes all the past history.
In the other direction, you can lose the Markov property by combining states, or "lumping". An example that I used in this MO answer, is to take a random walk $(S_n)$ on the integers, and define $Y_n=1[S_n>0]$. If there is a long string of time points with $Y_n=1$, then it is quite likely that the random walk is nowhere near zero and that the next value will also be 1. If you only know that the current value is 1, you are not as confident that the next value will be 1. Intuitively, this is why $Y_n$ doesn't have the Markov property.
For cases of lumping that preserve the Markov property, see this MSE answer.
Here are two examples, in discrete time and in continuous time, hopefully not too close from each other.
1. The process of the running maxima of a Markov process is in general not Markov.
1.1 Consider your favorite Markov process, say the standard symmetric random walk $(X_n)_{n\geqslant0}$ on the integer line, defined by $X_0=0$ and $X_n=Y_1+\cdots+Y_n$ for every $n\geqslant1$, where $(Y_n)_{n\geqslant1}$ is i.i.d. and symmetric Bernoulli, hence $\mathrm P(Y_n=1)=\mathrm P(Y_n=-1)=\frac12$. The process of the running maxima of $(X_n)_{n\geqslant0}$ is the process $(M_n)_{n\geqslant0}$ defined by $$M_n=\max\{X_k\,;\,0\leqslant k\leqslant n\}.$$ Then:
The process $(M_n)_{n\geqslant0}$ is not a Markov process, nor a Markov process of any higher order.
To see this, note that $M_{n+1}$ is either $M_n$ or $M_n+1$, and that the probability that $M_{n+1}=M_n+1$ depends on $$T_n=\max\{0\leqslant k\leqslant n\,;\,M_{n-k}=M_n\},$$ the time elapsed since the process $(X_n)_{n\geqslant0}$ last hit its current maximum $M_n$. Then, due to the symmetry of the increments of the random walk, $\mathrm P(M_{n+1}=M_n+1\,\mid\,\mathcal M_n)=u(T_n)$, where $\mathcal M_n=\sigma(M_k\,;\,0\leqslant k\leqslant n)$ and, for every $k\geqslant0$, $u(k)=\frac12\mathrm P(X_k=0)$. Thus, $u(2k-1)=0$ and $u(2k)=\frac12{2k\choose k}2^{-2k}$ for every $k\geqslant1$, hence the sequence $(u(2k))_{k\geqslant1}$ is decreasing. Since, for every $0\leqslant k\leqslant n$, $[T_n\geqslant k]=[M_{n-k}=M_n]$, this shows that the conditional probability that $[M_{n+1}=M_n+1]$ depends on the past in a possibly unlimited way.
1.2 The analogous continuous time process is the standard Brownian motion $(B_t)_{t\geqslant0}$. Consider $S_t=\sup\{B_s\,;\,0\leqslant s\leqslant t\}$. Then $(B_t)_{t\geqslant0}$ is a Markov process but $(S_t)_{t\geqslant0}$ is not.
2. Other examples without the Markov property are the processes of local times.
2.1 In the discrete setting, consider $$Z_n=\sum\limits_{k=1}^{n}[X_{2k}=0].$$ Then $Z_{n+1}$ is either $Z_n$ or $Z_n+1$, and the probability that $Z_{n+1}=Z_n+1$ depends on $$\max\{0\leqslant k\leqslant n\,;\,X_{2n-2k}=0\},$$ the time elapsed since the last zero of the random walk. For reasons similar to the ones explained for the maxima processes:
The process $(Z_n)_{n\geqslant0}$ is not a Markov process, nor a Markov process of any higher order.
2.2 The analogous process for the standard Brownian motion $(B_t)_{t\geqslant0}$ is the so-called local time at zero $(L_t)_{t\geqslant0}$. Likewise, $(L_t)_{t\geqslant0}$ is not a Markov process.
Swimming in a pool can be considered as a non-Markovian process. When the swimmer stops he feels a wave hitting him from behind after a short time. Of course the wave is created by the swimmer. This wave is the non-Markovian property. Instead of wave we can consider it as a mass of water following the swimmer.
Suppose that at a given instant t, a swimmer is swimming with a speed v. The amount of energy or power that he needs to maintain this speed depends on the past. If he was swimming faster (slower) just a second before, he will need less (more) energy at instant t to maintain his speed. Hence his situation at time t is effected by the past.