Where did we use induction process here? 
$(5.2)$ Proposition. Let $\varphi:F\to\tilde F$ be an isomorphism of fields. Let $f(x)$ be a nonconstant polynomial in $F[x]$, and let $\tilde f(x)$ be the corresponding polynomial in $\tilde F[x]$. Let $K$ and $\tilde K$ be splitting fields for $f(x)$ and $\tilde f(x)$. There is an isomorphism $\psi:K\to\tilde K$ which restricts to $\varphi$ on the subfield $F$ of $K$.



Theorem. Let $K$ be the splitting field of a polynomial $f(x)\in F[x]$. Then $K$ is a Galois extension of $F$; that is, $|G(K/F)|=[K:F]$.
We will prove the theorem by going back over the proof of Proposition $(5.2)$, keeping careful track of the number of choices.
$(5.4)$ Lemma. With the notation of $(5.2)$, the number of isomorphisms $\psi:K\to\tilde K$ extending $\varphi$ is equal to the degree $[K:F]$.
The theorem follows from this lemma if we set $\tilde F=F$, $\tilde K=K$, and $\varphi=\text{identity}$.$~\square$
$~~~~~$ Proof of Lemma $\it (5.4)$. We proceed as in the proof of Proposition $(5.2)$, choosing an irreducible factor $g(x)$ of $f(x)$ and one of the roots $\alpha$ of $g(x)$ in $K$. Let $F_1=F(\alpha)$. Any isomorphism $\psi:K\to\tilde K$ extending $\varphi$ will send $F_1$ to some subfield $\tilde F_1$ of $\tilde K$. This field $\tilde K$ will have the form $\tilde F(\tilde\alpha)$, where $\tilde\alpha=\psi(\alpha)$ is a root of $\tilde g(x)$ in $\tilde K$.
$~~~~~$ Conversely, to extend $\varphi$ to $\psi$, we may start by choosing any root $\tilde\alpha$ of $\tilde g(x)$ in $\tilde K$. We then extend $\varphi$ to a map $\varphi_1:F_1\to\tilde F_1=\tilde F(\tilde\alpha)$ by setting $\varphi_1(\alpha)=\tilde\alpha$. We use induction on $[K:F]$. Since $[K:F_1]<[K:F]$, the induction hypothesis tells us that for this particular choice of $\varphi_1$. there are $[K:F_1]$ extensions of $\varphi_1$ to an isomorphism $\psi:K\to\tilde K$. On the other hand, $\tilde g$ has distinct roots in $\tilde K$ because $g$ and $\tilde g$ are irreducible $[$Chapter $13~(5.8)]$. So the number of choices for $\tilde\alpha$ is the degree of $g$, which is $[F_1:F]$. There are $[F_1:F]$ choices for the isomorphism $\varphi_1$. This gives us a total of $[K:F_1][F_1:F]=[K:F]$ extensions of $\varphi$ to $\psi:K\to\tilde K$.$~\square$

This is from Artin Algebra. I get the proposition. I don’t get the induction part. Which case did we prove? and how does the rest of the induction process work? Sorry I don’t get the induction part.
Edit:
Can I just prove the case $[K:K]=1$? If so, why
 A: Let's write it differently and for simplicity assume $\tilde K$ is algebraically closed.
This is the induction step:
Whenever we find an intermediate field $F_1$, we have $$\tag1[K:F]=[K:F_1]\cdot[F_1:F]$$ with each factor $>1$ and hence $<[K:F]$. Then every homomorphim $\psi\colon K\to \tilde K$ with $\psi|_F=\phi$ gives us a homomorphism $\chi:=\psi|_{F_1}\colon F_1\to\tilde K$ with $\chi|_F=\phi$. By induction hypothesis (i.e., because the intermediate degrees are $<[K_G]$), we know that there are $[F_1:F]$ possible choices for $\chi$, and for each $\chi$, there are $[K:F_1]$ possible choices for $\psi$. Hence in total, there are $[K:F_1]\cdot [F_1:F]=[K:F]$ possible choices for $\psi$, i.e., $[K:F]$ homomoprhisms $K\to \tilde K$ that extend $\phi$.
The base case is when there is no intermediate field $F_1$ that allows us to have factors $<[K:F]$ in $(1)$. But then either $K=F$ and trivially the number of "extensions" is $1$, or $K=F[\alpha]$ for some (in fact, any) $\alpha\in K\setminus F$. The question becomes: How many homomorphisms $F(\alpha)\to\tilde K$ extend $\phi$? As $\alpha$ is root of an irreducible polynomials $f\in F[X]$, its image must be a root of the corresponding polynomial $\tilde f \in\tilde F[X]$. And in fact picking any such root as image of $\alpha$ does give us an extension of $\phi$ to $ F(\alpha)$. Hence the number of extensions equals the number of roots of $f$. In general it is false, but I assume that Artin considers only special cases (e.g., field of characteristic zero)  where in fact $[F(\alpha):F]=\deg f$.
