Prove that every polynomial in $\mathbb{R}[x]$ is continuous everywhere
Note: this proof should only use the epsilon-delta definition of a limit. Nonetheless, it's probably really easy, even if I mess up on a few steps.
Here's my attempt. Feel free to point out anything wrong/inefficient.
A polynomial $p(x)$ in $\mathbb{R}[x]$ is of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0.$
Lemma: If $f(x)$ and $g(x)$ are continuous at $a,$ then $f(x)+g(x)$ is continuous at $a.$
Proof of lemma: Since $f(x)$ is continuous at $a,$ $\forall \epsilon >0,\exists \delta >0 \; (0<|x-a|<\delta \Rightarrow |f(x)-f(a)|<\dfrac{\epsilon}{2}.$ Similarly, since $g(x)$ is continuous at $a,$ $\forall \epsilon >0,\exists \delta >0 \; (0<|x-a|<\delta \Rightarrow |g(x)-g(a)|<\dfrac{\epsilon}{2}.$
Thus we have that $|f(x)-g(x)-(f(a)-g(a))|\le |f(a)-f(a)|+|g(x)-g(a)|<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon.$ ${\square}$
Thus, from the lemma, by induction, if $a_ix^i$ is continuous at a point $a$ for some $i\in [0,n],$ then $p(x)$ is continuous. So it suffices to show that $a_ix^i, 0\le i \le n$ is continuous at a point $a.$ We need to estimate $|a_ix^i-a_ia^i|=|a_i||(x^i-a^i)|$ when $0<|x-a|<\delta.$ We know that $|x^i-a^i| =|x-a||x^{i-1}+ax^{i-2}+\dots+a^{i-1}|.$ Now, $|x^{i-1}+ax^{i-2}+\dots+a^{i-1}|$ is finite, so let $C$ denote its value. Choose $\delta = \dfrac{\epsilon}{Ca_i}$ and we have that $|a_ix^i-a_ia^i| <|a_i|\left(\dfrac{\epsilon}{Ca_i}C\right)=\epsilon,$ and so we are done.
Best Answer
Your proof in its broad strokes is good, though there is the one thing which I pointed out in the comments, regarding:
This $C$ is used to define $\epsilon$, and $\epsilon$ should be written not in terms of $x$, but in terms of only $\delta$ and $a$.
A more minor point, the pattern in the $+\dots+$ is not the clearest to see because the first and last terms both have the power $i+1$. Its easier (IMO) to read $\sum_{j=0}^{i-1} a^j x^{i-1-j}$.
Anyway, for the first point, you can note that $|x| \le |a| + \delta$. I suggest that you convince yourself that for $i>1$,
$$ \left| \sum_{j=0}^{i-1} a^j x^{i-1-j}\right| \le \sum_{j=0}^{i-1} |a|^j \big||a|+\delta\big|^{i-1-j}.$$ If you wanted something cleaner on the right, you could try $$ \left| \sum_{j=0}^{i-1} a^j x^{i-1-j}\right| \le i(|a|+\delta)^{i-1}.$$
Its not clear to me what you would mean by a "nonlinear combination of $x$ is finite"; for instance, $\frac1{x-a}$ is not bounded. So its better to just write the inequality you actually mean.
PS the "standard" proof (at least, whereever I looked) uses the following lemmas; if $f,g$ are continuous at $a$, and $c$ is a constant, then:
Now any polynomial is built by applying the first 3 operations on $ x$ and $1$, so the result follows.