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There is a definition:

Characteristic: the least common multiple (can be zero) of all orders of all elements in a field.

Then for a field with characteristic 0, pick elements,
$$ alpha_1, alpha_2, ..., alpha_n$$
such that the least common multiple of their orders is $q$.

Then the irreducible,
$$ f(x) = alpha_0 + alpha_1 x^q + alpha_2 x^{2q} + ... + alpha_n x^{nq} $$
has derivative,

$$ f'(x) = qalpha_1x^{q-1} + 2qalpha_2x^{2q-1} + ... + nqalpha_nx^{nq-1} = 0 $$

It follows that $gcd(f, f') = f$, and $f$ is inseparable.

Note that this can also work in field of characteristic $p$, if we choose such $q$ less than $p$.

So could anyone proof this is wrong, in the hope that there does not exist such a $q$?

How can any non-zero $a_i$ in a field of characteristic $0$ have finite additive order? Suppose $a$ has finite order so that $$0=a+a+ dots a=(1+1+dots +1)a$$

Now multiply by $a^{-1}$ to obtain $$0=1+1+ dots +1$$ with say $q$ summands.

Then multiply this by an arbitrary field element $b$ to show that $b$ has order at most $q$, and every element of the field has finite order which is a factor of $q$. And the characteristic is therefore non-zero.

As noted in comments, the definition you give of characteristic is deceptive. Actually, your argument is sound: indeed

if there exist elements $alpha_1, alpha_2, dots, alpha_n$ having finite order $>1$ such that the polynomial $$f(x) = alpha_0 + alpha_1 x^q + alpha_2 x^{2q} + dots + alpha_n x^{nq}$$ is irreducible, then $f$ is inseparable

is a true statement.

The only problem is that such elements don't exist.

The reason is very simple: in a field $F$ of characteristic $0$, there is no element $alphane0$ with finite order $q>0$. Indeed, if $qalpha=0$, with $alphane0$, then, for every $betain F$,
$$
qbeta=qalphaalpha^{-1}beta=0
$$
so the least common multiple of the orders of elements in $F$ is a divisor of $q$, hence it cannot be $0$: contradiction.

Similarly, if an element $alphane0$ in the field $F$ of characteritic $p>0$ has order $q$, then, for every element $betain F$,
$$
qbeta=qalphaalpha^{-1}beta=0
$$
and therefore the least common multiple of the orders of elements is a divisor of $q$. In particular $pmid q$. Thus you cannot find elements of order $q<p$.

In a field $F$, the characteristic is the order of the identity element (taking it to be $0$, if the identity element has infinite order). Indeed, if the order of the identity element is $p>0$, then for every $alphain F$ we have $palpha=p1alpha=0$, so the order of $alpha$ is a divisor of $p$. In the case of characteristic $0$, no nonzero element can have finite order, as shown before.

Over a ring (not a field), the argument above doesn't apply, but there is a much easier way to introduce the notion of characteristic: if $1$ is the identity element of the ring $R$, there exists a unique ring homomorphism $chicolonmathbb{Z}to R$, defined by $chi(n)=n1$ (the usual group-theoretic multiple).

If $kmathbb{Z}$ is the kernel of $chi$ (and $kge0$ is uniquely determined), then by definition of kernel, $kalpha=0$, for every $alphain R$. If $k>0$, then the least common multiple of the orders of elements is obviously $k$ (because one element with order $k$ exists, namely $1$). If $k=0$, then all elements except $0$ have infinite order.