Necessary and sufficient condition for real analyticity

We can do this with real analysis methods. WLOG $a=0.$ Suppose $f(x) = \sum_{k=0}^{\infty}c_kx^k,$ the series converging for $|x|<R.$

Claim: If $r<R/2,$ then $M_n(f,r)r^n/n! \to 0.$

Some preliminaries before proving the claim: The derivatives of any such $f$ are

$$\tag 1 f^{(n)}(x) = \sum_{k=n}^{\infty}\frac{k!}{(k-n)!}c_kx^{k-n},\,\, |x| < R.$$

Let's look at an important special case: $f(x) = 1/(1-x) = \sum_{k=0}^{\infty}x^k,$ where $R=1.$ Here we have

$$\tag 2f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}} =\sum_{k=n}^{\infty}\frac{k!}{(k-n)!}x^{k-n}.$$

For any $r<1,$ $M_n(f,r) = n!/(1-r)^{n+1}.$ Thus

$$\tag 3 \frac{M_n(f,r)r^n}{n!} = \frac{r^n}{(1-r)^{n+1}} = \frac{1}{1-r}\cdot \left ( \frac{r}{1-r} \right )^n.$$

Now if $r<1/2,$ then $r/(1-r) < 1,$ hence $(3)\to 0.$ So we have proved the claim in the special case of $f(x)=1/(1-x).$ This will help in proving the general case.

Proof of claim in general: We have $f(x) = \sum_{k=0}^{\infty}c_kx^k.$ Suppose $ r < R/2.$ Because $\sum_{k=0}^{\infty}|c_k|(R/2)^k < \infty,$ there is a constant $C$ such that $|c_k|(R/2)^k\le C$ for all $k.$ From $(1)$ we see

$$M_n(f,r) \le \sum_{k=n}^{\infty}\frac{k!}{(k-n)!}|c_k|r^{k-n}= \sum_{k=n}^{\infty}\frac{k!}{(k-n)!}|c_k|(R/2)^{k-n}(r/(R/2))^{k-n}$$ $$ = \frac{1}{(R/2)^n}\sum_{k=n}^{\infty}\frac{k!}{(k-n)!}|c_k|(R/2)^k(r/s)^{k-n} \le \frac{C}{(R/2)^n}\sum_{k=n}^{\infty}\frac{k!}{(k-n)!}(r/(R/2))^{k-n}$$ $$ =\frac{C}{(R/2)^n} \frac{n!}{(1-r/(R/2))^{n+1}}.$$

We have used $(2)$ to get the last expression. Thus

$$\tag 4\frac{M_n(f,r)r^n}{n!} \le C \frac{(r/(R/2))^n}{(1-r/(R/2))^{n+1}}.$$

Just as we saw with $(3),$ because $r/(R/2)<1,$ we have $(4)\to 0.$ This proves the claim.

Yes, and one way to show it is using a variant of Cauchy's estimate. If $R$ is less than the radius of convergence of the power series for $f$ at $a$, then $f$ extends to be analytic in a neighborhood of the closed disk $\{z\in\mathbb C:|z-a|\leq R\}$. For each $r$ with $0<r<R$, if $|z-a|\leq r$, then
$$\left\lvert f^{(n)}(z)\right\rvert \leqslant \frac{n! MR}{(R-r)^{n+1}},$$ where $M$ is a uniform bound for $f$ on $|z-a|\leq R$ (copied from Daniel Fischer's answer at the linked question). If $r\leq \frac12R$, then this implies that $M_n\frac{r^n}{n!}\leq \frac{M R}{R-r}<\infty$ for all $n$.

This is slightly stronger than what you ask because your $M_n$ from the real interval can be smaller than the bound in the disk. (I don't know of a way to show it without using complex analysis.)