There are two crucial results here.
Dedekind's theorem: Let $f$ be a monic irreducible polynomial over $\mathbb{Z}$ of degree $n$ and let $p$ be a prime such that $f$ has distinct roots $\bmod p$ (this is true for precisely the primes not dividing the discriminant). Suppose that the prime factorization of $f \bmod p$ is
$$f \equiv \prod_{i=1}^k f_i \bmod p.$$
Then the Galois group $G$ of $f$ contains an element of cycle type $(\deg f_1, \deg f_2, ...)$. In particular, if $f$ is irreducible $\bmod p$, then $G$ contains an $n$-cycle.
Frobenius density theorem: The density of the primes with respect to which the factorization of $f \bmod p$ has the above form is equal to the density of elements of $G$ with the corresponding cycle type. In particular, for every cycle type there is at least one such prime $p$.
It follows that
$f$ is reducible $\bmod p$ for all $p$ if and only if $G$ does not contain an $n$-cycle.
The smallest value of $n$ for which this is possible is $n = 4$, where the Galois group $V_4 \cong C_2 \times C_2$ has no $4$-cycle. Thus to write down a family of examples it suffices to write down a family of irreducible quartics with Galois group $V_4$. As discussed for example in this math.SE question, if
$$f(x) = (x^2 - a)^2 - b$$
is irreducible and $a^2 - b$ is a square, then $f$ has Galois group $V_4$. In particular, taking $b = a^2 - 1$ the problem reduces to finding infinitely many $a$ such that
$$f(x) = x^4 - 2ax^2 + 1$$
is irreducible. We get your examples by setting $a = 0, 5$.
By the rational root theorem, the only possible rational roots of $f$ are $\pm 1$, so by taking $a \neq 1$ we already guarantee that $f$ has no rational roots. If $f$ splits into two quadratic factors, then they both have constant term $\pm 1$, so we can write
$$x^4 - 2ax + 1 = (x^2 - bx \pm 1)(x^2 + bx \pm 1)$$
for some $b$. This gives
$$2a = b^2 \mp 2.$$
Thus $f$ is irreducible if and only if $2a$ cannot be written in the above form (and also $a \neq 1$).
Classifying polynomials $f$ with this property seems quite difficult in general. When $n = 4$, it turns out that $V_4$ is in fact the only transitive subgroup of $S_4$ not containing a $4$-cycle, but for higher values of $n$ there should be lots more, and then one has to tell whether a polynomial has one of these as a Galois group...
(Except if $n = q$ is prime; in this case $q | |G|$ so it must have a $q$-cycle.)
Degree $1$, clearly $x$ and $x+1$.
Degree $2$, notice the last coefficient must be one, so there are only two options, $x^2+x+1$ and $x^2+1$. Clearly only $x^2+x+1$ is irreducible.
Degree $4$. There are $8$ polynomials to consider, again, because the last coefficient is $1$, now notice a polynomial is divisible by $x+1$ if and only if the sum of its coefficients is even. So the only polynomials without factors of degree $1$ are four:
$x^4+x^3+x^2+x+1$
$x^4+x^3+1$
$x^4+x^2+1$
$x^4+x+1$.
Of course, we are missing the possibility it is the product of two irreducibles of degree $2$, but the only combination is $(x^2+x+1)(x^2+x+1)=x^4+x^2+1$.
Hence the irreducible ones are:
$x^4+x^3+x^2+x+1,x^4+x^3+1,x^4+x+1$
Best Answer
Even if you don't know the explicit factorization, you certainly know that a quartic polynomial in $\mathbb{R}[X]$ is reducible.
Indeed, a consequence of the fact that every polynomial of positive degree in $\mathbb{C}[X]$ has a root, we can see that the irreducible polynomials over $\mathbb{R}[X]$ are those of degree $1$ and the quadratic polynomials with negative discriminant.
To wit, let $f(X)\in\mathbb{R}[X]$ be monic of degree $>1$. If $f$ has a real root, then it is reducible. So, assume it has no real root. Since it has a complex root $a$, also $\bar{a}$ is a root, because $$ f(\bar{a})=\overline{f(a)}=\bar{0}=0 $$ (overlining means conjugation). Thus $f$ is divisible (in $\mathbb{C}$) by $X-a$ and $X-\bar{a}$, so also by their product $$ g(X)=(X-a)(X-\bar{a})=X^2-(a+\bar{a})X+a\bar{a} $$ (because $X-a$ and $X-\bar{a}$ are coprime in $\mathbb{C}[X]$). However, $g(X)$ has real coefficients, so the quotient of the division of $f(X)$ by $g(X)$ is again a polynomial in $\mathbb{R}[X]$. So, unless $f$ has degree $2$ (and so $f(X)=g(X)$), $f$ is reducible.