The revised question, where you just want uniformly bounded Holder norm on the approximating polynomials, the answer is **yes**. 

In fact, you have more. 

**Theorem 1** If $f\in C^{m,\alpha}_M([0,1])$ is such that the function $\tilde{f}(x) = \begin{cases} f(x) & x\in [0,1]\\  0 & x\not\in [0,1]\end{cases}$ is in $C^{m,\alpha}(\mathbb{R})$, then there exists a polynomial approximation sequence in $C^{m,\alpha}_M([0,1])$. 

This theorem is based on a proof of the Stone-Weierstrass Theorem using a polynomial approximation to identity. See e.g. Theorem 11.7.1 of [Lebl's textbook (volume 2)](https://www.jirka.org/ra/). 

A nice property of the convolution argument is that:

_If $\phi$ is a non-negative function with total integral 1, then $\|\phi*f\|_{\infty} \leq \|f\|_\infty$._

Since $(\phi*g)^{(j)} = \phi* (g^{(j)})$, for any function $g$, applying this to $\tilde{f}$ you see that the convolution by the polynomial approximation to identity means that the polynomials constructed have $C^m$ norms bounded by that of $\tilde{f}$. 

Similarly, letting $g = \tilde{f}^{(m)}$, observe for each fixed $t\neq 0$, $x\mapsto \frac{1}{|t|^\alpha}({g}(x+t) - {g}(x))$ is a $C^{0,\alpha}$ function, we have that 

$$ \frac{1}{|t|^\alpha} \| \phi*{g}(x+t) - \phi*{g}(x)\|_{L^\infty_x} \leq \frac{1}{|t|^\alpha} \|{g}(x+t) - {g}(x)\|_{L^\infty_x} $$

Taking the sup of both sides over $t\neq 0$ this shows that the Holder semi-norm of $\phi*\tilde{f}$ is bounded by that of $\tilde{f}$. And this proves the Theorem. 

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Now, given $f\in C^{m,\alpha}_M([0,1])$, there exists a polynomial $q$ such that $f-q$ satisfies the hypotheses of Theorem 1. The polynomial $q$ is determined entirely by $f(0), f'(0), \ldots, f^{(m)}(0), f(1), f'(1), \ldots, f^{(m)}(1)$.)  You can select $q$ to be of degree $2m+1$, in which case the $q$ is uniquely determined by those values. The relation between those values and the coefficients of $q$ are given by a linear operator, depending only on $m$. This means that there exists $\tilde{M}$ that uniformly bounds the coefficients of $q$, where $\tilde{M}$ depends only on $M$ and on $m$ (since by assumption $f(0), \ldots f^{(m)}(1)$ are all bounded by $M$). (Note that in the context of Theorem 1, this unique $q$ is the 0 polynomial, and hence $\tilde{M} = 0$.) The mapping from $\mathbb{R}^{2m+2}\times \mathbb{R} \to \mathbb{R}$ where the first factor represents the coefficients of $q$ and the last factor represents $x$ mapping $(q,x)\mapsto q(x)$ is continuous (in fact a polynomial), and hence achieves a maximum on compact regions. Therefore knowing that the coefficients are bounded by $\tilde{M}$ and $x\in [0,1]$ means that there exists $\hat{M}$ depending only on $M$ and $m$ that bounds the $C^{m+1}$ and hence $C^{m,\alpha}$ norm of $q$. Finally set $M' = M + 2 \hat{M}$. 

So you conclude:

**Theorem 2** Given $m, \alpha,M$, there exists $M' = M'(m,M,\alpha)$ such that whenever $f\in C^{m,\alpha}_M([0,1])$, you can approximate $f$ by polynomials $p_j$ in the uniform norm, with $p_j$ selected from $C^{m,\alpha}_{M'}([0,1])$.