If $K$ is a field with an infinite number of elements and $f(x),g(x)\in K[x]$ such that for $\forall \alpha\in K$ $f(\alpha)=g(\alpha)$, then $f(x)=g(x)$.
My attempt at the proof (by contradiction):
Lets assume that $f(x)\neq g(x)$, and without loss of generality that $deg(f)\geq deg(g)$, where $deg(f)=n$, $deg(g)=m$.
Then $h(x)=f(x)-g(x)\neq 0$, and $0<deg(h)\leq n$. (If $deg(h)=0$ then $f(\alpha)-g(\alpha)=c\in K$ and $f(\alpha)\neq g(\alpha)$)
Therefore $h(x)$ has $k$ roots ${\alpha_1,\dots,\alpha_k}$ where $k\leq n$
Now Let $\beta_1,\dots,\beta_{n-k+1}\in K$, such that $\beta_j\neq\alpha_i$ for $i=1,\dots,n$ and $j=1,\dots,n-k+1$
Since $f(\beta_j)=g(\beta_j)$, $h(\beta_j)=0$. $\beta_1,\dots,\beta_{n-k+1}$ are also roots of $h(x)$ and $h(x)$ has more than $n$ roots !
Therefore $f(x)=g(x)$
The proof is fine. But there are a couple of aesthetic concerns.
I would say $h=f-g\neq 0$, not $h(x)=f(x)-g(x)\neq0$. I know you have established that $x$ is the variable in the polynomial ring, but to me it looks a bit too much like evaluation. And when evaluating, they are equal to $0$.
There is no need for that much contradiction, and no need for the $\alpha$ and $\beta$. Just take $f$ and $g$ to have all (in particular infinitely many) values in common. Then $ h$ has all points (in particular infinitely many points) as roots.
At this stage, if you're not comfortable with concluding immediately that $h=0$, you can do contradiction (presumably by induction on the degree of $h$ or something similar) to prove that $h=0$.