From Wikipedia I learn that finite fields are not closed: If $F$ has $a_1,\dots,a_n$ then one could construct a polynomial $f(x) = (x-a_1)\cdot \ldots \cdot (x-a_n) + 1$ which has no roots (no zeros in $F$).
However, we don't need to go that far in creating the polynomial that uses every root, shouldn't $f(x) = x^2 + 1$ show that it has no roots? Am I right or am I missing something fundamental?
Both you and Wikipedia are right. Let me elaborate. Wikipedia tries to give a proof that economizes on assumptions. That is, the polynomial they construct is by a uniform procedure and works for all finite fields and the proof is just one line.
In contrast in your method you are trying to economize on the degree of the polynomial. So you want to give a polynomial that does not have a root, degree being 2 irrespective of the field, while Wikipedia polynomial has degree as big as the cardinality of the finite field.
Existence of a solution to the equation $x^2+1=0$ is equivalent to finding an element of order $4$ in the multiplicative group of the field. So using Lagrange theorem we can show this polynomial works when the order of the field is NOT 1 modulo 4. For fields where this hypothesis is not valid you need to produce a different polynomial.
Remember Euclid's proof of the infinitude of prime numbers: the prime that is shown to exist is not necessarily the immediate next. (If $2,3,5$ are the only prime numbers we know then $2\times3\times5 +1= 31$ is not first prime after $5$).