To prove it rigorously, for example, the Markov property for the Brownian Bridge, we have to do some calculus. Let $(W_t)_{t \geq 0}$ be the standard Brownian motion issued from $0$, and we have $(B_t)_{s \leq t \leq T}$ defined as
$$ B_t = x + W_t - W_s - \frac{t-s}{T-s}(x + W_T -W_s) ,$$
a Brownian Bridge between $s, T$ and at $s$ it is $x$ and at $T$ its value is $0$. One way to see this formula is that the term $x + W_t - W_s$ is the Brownian Motion while we have to reduce the term at the endpoint $T$. Some simple calculus shows that it is equal to
$$B_t = \frac{T-t}{T-s}(x + W_t - W_s) - \frac{t-s}{T-s}(W_T - W_t).$$
A Markov property is very simple but requires calculus: we would like to show that for $s < r < t < T$ we have
$$B_t = B_r + W_t - W_r - \frac{t-r}{T-r}(B_r + W_T -W_r). \quad (\star)$$
Now we prove it. An intermediate step tells us
$$ - \frac{t-r}{T-r}(B_r + W_T -W_r) = - \frac{t-r}{T-s}(x + W_T - W_s)$$ and we put it into the formula that
$$RHS (\star) = x + W_r - W_s - \frac{r-s}{T-s}(x + W_T -W_s) + W_t - W_r - \frac{t-r}{T-s}(x + W_T - W_s)\\ = x + W_t - W_s - \frac{r-s}{T-s}(x + W_T -W_s) - \frac{t-r}{T-s}(x + W_T - W_s) \\ = x + W_t - W_s - \frac{t-s}{T-s}(x + W_T -W_s).$$
This proves the Markov property. Then the strong Markov property is just an approximation and the regularity of the trajectory.
