How is this induction argument here valid I have this proof of 'existence of splitting fields' from Gallian's Contemporary Abstract Algebra (I have seen it elsewhere as well, and it goes in similar lines):

Let $F$ be a field and let $f(x)$ be a non-constant element of $F[x]$. Then there exists a splitting field $E$ for $f(x)$ over $F$.
Proof: We proceed by induction on $\deg f(x)$. If $\deg f(x) = 1$, then $f(x)$ is linear. Now suppose that the statement is true for all fields and all polynomials of degree less than that of $f(x)$. Then, there is an extension $E$ of $F$ in which $f(x)$ has a zero, say, $a_1$ . Then we may write $f(x) = (x - a_1 )g(x)$, where $g(x) \in E[x]$. Since $\deg g(x) < \deg f(x)$, by induction, there is a field $K$ that contains $E$ and all the zeros of $g(x)$, say, $a_2 , . . . , a_n$ . Clearly, then, a splitting field for $f(x)$ over $F$ is $F(a_ 1 , a_ 2 , . . . , a_ n )$.

I can't digest how can we, in the induction step, assume that 'the statement is true for all fields '. I have seen this kind of argument for the first time. Please help me!

The full statement of the induction hypothesis is 

The theorem is true for all fields $F$ and for all non-constant elements of $F[x]$ having degree $\le n$. 

You'll find that the induction hypothesis is often not spelled out in full formal detail like that. Instead it becomes the reader's responsibility to figure out the appropriate formal statement of the induction hypothesis.
