Let $f(x)\in\mathbb{Q}[x]$ be an irreducible polynomial of degree $n$, whose zeros are all real. If $K$ is a splitting field of $f(x)$, it is true that $\mathrm{Gal}(K/\mathbb{Q})$ can be embedded into $A_{n}$? If $f(x)$ has complex zeros, then its Galois group has a complex conjugation map which corresponds to a transposition in $S_{n}$. Hence I thought that Galois group of polynomial with real zeros will have no odd permutation, which implies that its Galois group is a subgroup of $A_{n}$.
I hope that this is true since this may give a new proof of inverse Galois problem for $A_{n}$. We can prove that $p(x) = A(x-1)\cdots (x-n)+1$ is irreducible and only has real zeros for some large $A$, so it would have a Galois group which is a subgroup of $A_{n}$ (if the above statement is true) and it seems that the Galois group will be $A_{n}$ for suitable choice of $A$, although I don't know whether this argument is true or not.
I guess in your notation you have that $n = \deg f$. If this is the case, then your assumption isn't right. For example take the polynomial:
$$p(x) = x^4 - 5x^3 + 3x^2 + 3x - 1$$
which is irreducible over $\mathbb{Q}$ and also has $4$ real roots. On the other side PARI/GP gives that the Galois group of $p(x)$ over $\mathbb{Q}$ is $S_4$ and from here it's obvious that it can't be embedded into $A_4$.
We can also calculate the Galois group of $p(x)$ over $\mathbb{Q}$ by using the method described in here. It's not hard to find out the cubic resolvent, which is:
$$R_3(x) = x^3 - 3x^2 - 11x + 4$$
This is an irreducible polynomial in $\mathbb{Q}[x]$. Combining it with the fact that $\operatorname{disc} f = 8789 = 11 \cdot 17 \cdot 47$, a non-square we conclude that the Galosi group of $p(x)$ over $\mathbb{Q}$ is isomorphic to $S_4$