If $q(X)$ is reducible in $\mathbb Z[X]$, then it's reducible in $\mathbb Z_p[X]$ for every prime $p$ My book states, without a proof, that

If $q(X)$ is reducible in $\mathbb Z[X]$, then it's reducible in $\mathbb Z_p[X]$ for every prime $p$.

The contrapositive of the above result is more useful: if one finds a prime $p$ such that $q(X)$ is irreducible in $\mathbb Z_p[X]$, then it's irreducible in $\mathbb Z[X]$.
I was wondering how one might prove the result, but then read this question:
Irreducible polynomial which is reducible modulo every prime,
which seems to be in contrast with the above statement. How is that possible?
 A: The following implication is true (as you mention):
$$q(X) \in \Bbb Z[X] \text{ monic reducible } \implies \bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ reducible for all prime } p$$
The contrapositive of this implication is:
$$\bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ irreducible for some prime } p \implies q(X) \in \Bbb Z[X] \text{ irreducible (or non monic) }$$
However, the implication
$$q(X) \in \Bbb Z[X] \text{ irreducible monic } \implies \bar q(X) \in \Bbb Z/p\Bbb Z[X] \text{ irreducible for some prime } p$$
is the converse of the first implication, which doesn't hold in general (as you provide a link to give a counter-example).

The first implication can be proven as follows:
if you assume $q= fg$ with $f,g \in \Bbb Z[X] \setminus \{±1\}$, then $\bar q = \bar f \bar g \in \Bbb Z/p\Bbb Z[X]$. If you suppose that the dominant coefficient of $q$ is not a multiple of $p$ (e.g. $q$ monic), then both $\bar f$ and  $\bar g$ have positive degree, hence $\bar q $ is reducible.

More generally, if $A$ is a UFD, $a \in A$ is an element, $f \in A[X]$ is a polynomial with content $1$ and with dominant coefficient invertible in $A/(a)$, then the irreducibility of $\bar f$ in $A/(a)[X]$ gives the irreducibility of $f$ in $A[X]$. [Berhuy, theorem IV.2.3].
A: The idea of the proof
Suppose $q(X)$ reducible over $\mathbb Z[X]$. Then, $$q(X)=R(X)P(X)$$
where $\deg q,\deg R>0$. In particular, in $\mathbb Z_p[X]$, $$\bar q(X)=\bar R(X)\bar P(X).$$
Now, you have to manage with primitive polynomials, since in $\mathbb Z_p[X]$, some polynomial can be null (that is not possible with primitive roots) (for example $2X^2+2$ in $\mathbb Z_2[X]$. Be careful, since $2$ is not a unit in $\mathbb Z$, you have that $2X^2+2$ is a reducible over $\mathbb Z$).
