polynomial converges uniformily to analytic function

Assume that $f \in C(\bar D) \cap \mathcal{O}(D)$. Then the Maclaurin polynomials of $f$ converge locally uniformly on $D$ to $f$, but in general it is not true that the Maclaurin polynomials converge uniformly to $f$ on $\bar D$. In fact, it is even possible to construct examples of such $f$:s where the Maclaurin series diverges at some points of $\partial D$. (See this question.)

If you really want uniform convergence on the whole of $\bar D$, you need to work a little more. Let $r_n$ be an increasing sequence of positive numbers tending to $1$ and let $$f_n(z) = f(r_n z).$$ Then each $f_n$ is analytic on a neighbourhood of $\bar D$.

Let $P_n$ be the $d_n$:th Maclaurin polynomial of $f_n$, where $d_n$ is chosen so large that $$\sup_{z\in\bar D} |P_n(z) - f_n(z)| < 2^{-n}$$ (This is possible, since the Maclaurin polynomials converge locally uniformly and $f_n$ is analytic on a neighbourhood of $\bar D$.)

Hence \begin{align} \sup_{z\in \bar D} |P_n(z) - f(z)| &\le \sup_{z\in \bar D} |P_n(z) - f_n(z)| + \sup_{z\in \bar D} |f_n(z) - f(z)| \\[10pt] &\le 2^{-n} + \sup_{z\in \bar D} |f_n(z) - f(z)|. \end{align} By uniform continuity of $f$ on $\bar D$, the second term tends to $0$ as $n\to\infty$ and we are done.

All of this is a (simple) special case of a deep theorem by Mergelyan: If $K$ is a compact subset of $\mathbb{C}$ with connected complement, and $f$ is continuous on $K$, analytic on the interior of $K$, then $f$ can be approximated uniformly on $K$ by polynomials.