In my topology class, I have learned that we can define a topological basis on the set of continuous functions $C[0,1]$ as
$$ \beta(a,t,\epsilon) = \{ f \in C[0,1] : |f(t) - a| < \epsilon \}. $$
In this sense, the topology induced is the pointwise convergence topology, meaning that any convergent sequences in the topology are exactly sequences of functions that convergences at a point.
Then I wondered about whether it is possible to write down the topological basis or topology whose convergence mode is uniform convergence. Specifically, I am already given the metric corresponding to uniform convergence, i.e. the $L_\infty$ metric:
$$ d_\infty(f,g) = \sup_{x \in [0,1]} |f(x) - g(x)|. $$
So it remains to construct base elements using the metric. By Weierstrass Approximation Theorem we also know that for any $ f \in C[0,1]$ and any positive real number $\epsilon$, there exists a polynomial $P$ such that
$$ d_\infty(f,P) < \epsilon. $$
From this, I define a set $ B(P,\epsilon) $ as
$$ B(P,\epsilon) := \{ f \in C[0,1]: P \text{ is a polynomial}, d_\infty(f,P) < \epsilon \} $$
Then I consider the collection of all $ B(P, \epsilon)$ in the sense that I consider all possible polynomials $P$ and all positive real number $\epsilon$, and call the collection $\mathcal{B}$.
In this sense, is $\mathcal{B}$ a topological basis for $C[0,1]$? If it is, does it induce the topology in which the mode of convergence is uniform convergence?