Let $f : \mathbb{R} \to \mathbb{R}$ be approximated arbitrarily well by polynomials of bounded degree. Prove $f$ is a polynomial.

For each $n$ you can find some $$P_n(x)=a_d^nx^d+...+a_1^nx+a_0^n$$ such that $$\sup_{x \in \mathbb{R}} |f(x) - P_n(x)| < \frac{1}{n}$$

Then, for each $n,m$ you have $$\sup_{x \in \mathbb{R}} |P_m(x) - P_n(x)| \leq \sup_{x \in \mathbb{R}} |f(x) - P_n(x)| +\sup_{x \in \mathbb{R}} |f(x) - P_m(x)| < \frac{1}{n}+ \frac{1}{m}$$

This shows that $P_m(x)-P_n(x)$ is a polynomial which is bounded on $\mathbb R$ and hence a constant. Therefore, there exists some polynomial $P(x)$ and some $c_n \in \mathbb R$ such that $$P_n(x)=P(x)+c_n$$

By the above you get $$\left| c_n -c_m \right| =\sup_{x \in \mathbb{R}} |P_m(x) - P_n(x)| < \frac{1}{n}+ \frac{1}{m} $$

Therefore, $c_n$ is Cauchy and hence convergent to some $c$.

We claim that $f(x)=P(x)+c$.

Let $\epsilon >0$. Pick some $N$ such that, for all $n >N$ we have $$\frac{1}{n} < \frac{\epsilon}{2}\\ |c_n-c|< \frac{\epsilon}{2}$$

Let $n >N$ be fixed but arbitrary. Then $$\sup_{x \in \mathbb{R}} |f(x) - P(x)| \leq \sup_{x \in \mathbb{R}} |f(x) - P_n(x)| +\sup_{x \in \mathbb{R}} |P_n(x) - P(x)| < \frac{1}{n}+ |c_n-c| <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

This shows that $$\sup_{x \in \mathbb{R}} |f(x) - P(x)| < \epsilon$$ for all $\epsilon >0$.

First pick a polynomial $g(x)$ for $\varepsilon = 1$, i.e., $\sup\limits_{x\in \mathbb{R}}|f(x)-g(x)|< 1$. Let $h(x)= f(x)-g(x)$. Then the same condition holds for $h(x)$: To obtain an approximation with error $\varepsilon$, find one for $f(x)$ and subtract $g(x)$.

So it is enough to verify the assertion for bounded functions. But that is trivial: a bounded function with this property must be constant. Indirect proof: if it had two different values (say at points $x,y$), let $\delta=|h(x)-h(y)|$, and put $\varepsilon:=\delta/3$. The approximating polynomial must be constant (otherwise we have a problem with the limit at infinity), but no constant is close enough at both $x$ and $y$. Hence, $h(x)$ is a constant, and then $f(x)= g(x)+h(x)$ is a polynomial.

Since $f$ is a uniform limit of polynomials (contiunous functions), in particular it is continuous.

Consider the space $C(\mathbb{R})$ of continuous functions $\mathbb{R} \to \mathbb{R}$ and equip it with a family of seminorms $\|\cdot\|_{\infty, [a,b]}$ given by $$\|g\|_{\infty, [a,b]} = \sup_{x\in[a,b]}|g(x)|$$

for all segments $[a,b] \subseteq \mathbb{R}$.

This turns $C(\mathbb{R})$ into a locally convex topological vector space.

Assume that $f$ can be uniformly approximated by polynomials $(p_k)_k$ of degree $\le n$.

The subspace $\mathbb{R}_{\le n}[x]$ of real polynomials of degree $\le n$ is a finite dimensional subspace of $C(\mathbb{R})$ and hence it is closed in $C(\mathbb{R})$. Since $p_k \to f$ uniformly on $\mathbb{R}$, in particular $\|f -p_k\|_{\infty, [a,b]} \to 0$ for all segments $[a,b] \subseteq \mathbb{R}$.

Hence $p_k \to f$ in the above topology so $f$ is in the closure of $\mathbb{R}_{\le n}[x]$. Since the subspace is closed, we conclude $f \in \mathbb{R}_{\le n}[x]$.