The point is this: if $R \to S$ is a homomorphism and the image of $f$ is irreducible in $S[x]$, then $f$ must also be irreducible in $R[x]$ (since any factorization of $f$ in $R[x]$ maps to a factorization of $f$ in $S[x]$). This idea gets to the heart of what the point of homomorphisms is: they are maps that send algebraic statements ($f$ has a factorization) to other algebraic statements.
I guess the best motivated way to approach this is to make two choices for $S$. First we choose $S = \mathbb{Z}/p\mathbb{Z}$. This is a surprisingly useful choice for specific polynomials, especially of low degree (it works for some prime $p$ if and only if the Galois group of the polynomial $f$ contains a $\deg f$-cycle, which always happens in degree $2, 3$ and is still very likely in higher degrees). For example the polynomial $x^3 - x + 1$ must be irreducible because it has no roots $\bmod 2$.
At some point we might encounter a polynomial like $x^3 - 2x - 2$. Now this polynomial is reducible $\bmod 2$ since it factors as $x^3 = x \cdot x \cdot x$, but this implies that if $x^3 - 2x - 2$ had an actual factorization $fg$ in the integers then $f, g$ would both have to have all their non-leading terms divisible by $2$, hence the constant term of $fg$ is divisible by $4$; contradiction. Eisenstein's criterion is a simple generalization of this example; it corresponds to choosing $S = \mathbb{Z}/p^2\mathbb{Z}$ or $S = \mathbb{Z}_p$.
Nowadays it is understood that Eisenstein's criterion is really a statement about total ramification, but my guess is that this was not known when it was first written down. My guess is that Eisenstein wrote down this criterion for the sole purpose of proving that the cyclotomic polynomials $\Phi_p(x) = x^{p-1} + ... + 1$ are irreducible, and here it is very natural to think about what happens $\bmod p$ since $(x - 1) \Phi_p(x) = x^p - 1$ splits into linear factors $\bmod p$ by Fermat's little theorem. It is not a big leap from here to the general Eisenstein criterion.
Suppose by contradiction that
$$\sqrt{2}=a+b\sqrt{3}+c\sqrt{5}+d \sqrt{15} \,.$$
Squaring both sides and using the fact that $1, \sqrt{3}, \sqrt{5}, \sqrt{15}$ are linearly independent over Q you get
$$2=a^2+3b^2+5c^2+15d^2 \,.$$
$$ab+5cd =0 \,.$$
$$ac+3bd=0 \,.$$
$$ad+bc=0 \,.$$
From the last two equations we get
$$3bd^2=-acd=bc^2 \,.$$
Since $3d^2=c^2$ has no rational roots we get that $b=0$.
It follows that
$$2=a^2+5c^2+15d^2 \,.$$
$$5cd =0 \,.$$
$$ac=0 \,.$$
$$ad=0 \,.$$
From the last two equations it follows that two of $a,c,d$ must be $0$, and then you get from the first equation that $x^2 \in \{ 2, \frac{2}{5}, \frac{2}{15} \}$ where $x$ is the nonzero one... Contradiction with $x \in Q$.
Best Answer
Let $\alpha \in K - \mathbb{Q}$. Since $1 , α, α^2$ are linearly dependent over $\mathbb{Q}$, $aα^2 + bα + c = 0$, where $a, b, c \in \mathbb{Q}$ and not all of $a, b, c$ are zero. By multiplying a suitable nonzero integer, we can assume $a, b, c \in \mathbb{Z}$. If $a = 0$, we get $bα + c = 0$. Since $b$ or $c$ is not zero, this can't happen. Hence $a \neq 0$. Hence $\mathbb{Q}(\alpha) = \mathbb{Q}((b^2 - 4ac)^{1/2})$. Since 1$ , α$ are linearly independent over $\mathbb{Q}$, [$\mathbb{Q}(\alpha) : \mathbb{Q}$] = 2. Hence $K = \mathbb{Q}(\alpha)$ and we are done.