Probability of "union" of events Let $\{X(t)\}$ be a stochastic process with $t\in [0, T]$. 
It is possible to prove that
$P(\exists t\in [0, T] : X(t) >a) \leq\int_0^TP(X(t)>a)dt$? 
My idea was to write the left hand side as union of events, the problem is this union is not countable... Is that a problem?  
 A: Since your stochastic process is completely arbitrary, it and the value $a$ are a distraction. We may as well just take a family $A_t$ of events indexed by $t\in[0, T]$ and ask if
$$P\left(\bigcup_t A_t\right)\leq \int_0^TP(A_t)dt$$
This is true in the discrete case. When you generalize the identity $P(\bigcup_n A_n)\leq \sum_n P(A_n)$ to a continuous index, there arises the question of what measure to use for the integral that's going to generalize that sum over $n$. In this case the choice of the Lebesgue measure is arbitrary. Since there are no conditions imposed on the events $A_t$, why should the Lebesgue measure be relevant here? Thus there's really no reason for this proposition to be true. And indeed, we can just let $P(A_0)=1$ and $P(A_t)=0$ for $t>0$.
In fact, there's a sense in which we can show that the inequality that holds in the discrete case cannot be generalized. The minimum we might ask of such a generalization would be that the right hand side - the thing that's going to replace $\sum P(A_n)$ - should only depend on the function $f(t)=P(A_t)$. It should only depend on the probabilities, and not on the set theoretic properties of the events $A_t$. After all, if it were allowed to depend on the events themselves, it could just be $P(\bigcup A_t)$.
If $f$ is of countable support, we're reduced to the discrete case. And if $f$ has uncountable support, then no such inequality is possible in general, other than the trivial bound $1$, since for any $f$ of uncountable support we can find families of events $A_t$ such that $P(\bigcup A_t)=1$. For example, we can take as probability space the interval $[0, 1]$ under the Lebesgue measure. If $f(t)>0$ for uncountably many $t$, then for some $n$, $f(t)>\frac 1 n$ for infinitely many $t$, and in particular for $n$ values of $t$, say $t_1, ... t_n$. Thus we can define the $A_{t_i}$ so as to cover of the unit interval, and we'll already have $P(\bigcup A_t)=1$. Define all the other $A_t$ however you like that's consistent with $f$, for example $A_t=[0, f(t)]$.
