The "usual" metric on $C([a,b], \mathbb{R})$ is
$$d(f,g) = \sup_{x \in [a,b]} |f(x) - g(x)|$$
Please let me know if that is not the metric you are assuming.
Let $P$ denote the set of polynomials in $C([a,b], \mathbb{R})$.
If $P$ is open, then for any function $f \in P$, there must be some $\epsilon$-neighborhood of $f$ which contains only elements of $P$. In other words, there must be some $\epsilon > 0$ such that, if $g \in C([a,b], \mathbb{R})$ and $d(f,g) < \epsilon$, then $g$ is a polynomial.
We will show that $P$ is not open by showing that for any $f \in P$ and any $\epsilon > 0$, we can find a non-polynomial in $C([a,b], \mathbb{R})$ for which $d(f,g) < \epsilon$. This will show that $f$ is not an interior point of $P$.
Let $f \in P$ and let $\epsilon > 0$. Then define $g(x) = f(x) + (\epsilon/2) \sin(x)$. Note that $g \in C([a,b], \mathbb{R})$ since $g$ is the sum of two continuous functions. Also note, that $g$ is not a polynomial: if it were, then $(\epsilon/2) \sin(x) = g(x) - f(x)$ would be the difference of two polynomials, hence a polynomial. But this is not the case, because $\sin(x)$ is not the zero function and has infinitely many zeros, whereas a nonzero polynomial can only have finitely many zeros.
Now we compute the distance between $f$ and $g$:
$$\begin{align}
d(f,g) &= \sup_x |g(x) - f(x)|\\
&= \sup_x |(\epsilon/2) \sin(x)|\\
&= (\epsilon/2) \sup_x |\sin(x)|\\
&= \epsilon/2 \end{align}$$
Therefore, $d(f,g) < \epsilon$ and we have found the desired contradiction.
Note that this proof shows that $P$ contains no interior points, which is a stronger statement than "$P$ is not open", which only requires us to show that $P$ contains at least one non-interior point.
Best Answer
A set $U\subset \mathbb R$ is open if and only if for every $x\in U$, there exists some $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon)$ is a subset of $U$.
For $U=\mathbb Z$, this is clearly not the case: