The following extension of Keivan Karai's comment proves the result.

Consider $q(x):=p(x)-c$ where $c=\frac12 \min p([0,1])$.

Approximate $q(x)$ as Keivan suggested: let $$ q_n(x) = \sum_{i=1}^n q(i/n) \binom ni x^i(1-x)^i . $$ Then $q_n$ is a polynomial of at most the same degree as $q$ (and hence $p$) and $q_n$ converges to $q$ uniformly on $[0,1]$. Since the degree is bounded, it follows that the coefficients of $Q_n$ converge to the coefficients of $q$. In particular, $\deg q_n=\deg q$ for all large enough $n$. Let $a$ be the leading coefficient of $q$ and $a_n$ the leading coefficient of $q_n$. Then $a_n\to a$ and hence the polynomial $$ r_n(x):=\frac a{a_n} q_n(x) $$ converges to $q(x)$ uniformly on $[0,1]$. Choose $n$ so large that $|r_n-q|<c$ on $[0,1]$, then $r_n<p$ on $[0,1]$. Since $r_n$ is already represented in the desired form, it remains to represent $p-r_n$. Observe that $\deg(p-r_n)<\deg p$ and proceed by induction.

Appendix: Why $\deg q_n\le\deg q$.

Since $q_n$ is linear in $q$, it suffices to consider the case $q(x)=x^k$. Let $$ F_0(u,v) = (u+v)^n = \sum_{i=0}^n \binom ni u^i v^{n-i} $$ and $$ F_m = u\cdot \frac{\partial}{\partial u} F_{m-1}, \qquad m=1,2,\dots $$ Then $$ F_k(u,v) = \sum_{i=1}^n i^k \binom in u^iv^{n-i}, $$ hence $q_n(x) = n^{-n} F_k(x,1-x)$. By induction in $k$ one sees that $F_k(u,v)$ is a linear combination of terms of the form $u^j(u+v)^{n-j}$ with $j\le k$. Hence $F_k(x,1-x)$ has degree at most $k$, and so does $q_n(x)$.

Remark. Analyzing the coefficients of the above linear combination, one can see that, as $n\to\infty$ and $k$ fixed, the coefficient at $u^j(u+v)^{n-j}$ goes to 1 and all other ones go to 0. This gives an elementary proof of convergence $q_n\to q$ (coefficient-wise) avoiding the general theory.

The following extension of Keivan Karai's comment proves the result.

Consider $q(x):=p(x)-c$ where $c=\frac12 \min p([0,1])$.

Approximate $q(x)$ as Keivan suggested: let $$ q_n(x) = \sum_{i=0}^n q(i/n) \binom ni x^i(1-x)^{n-i} . $$ Then $q_n$ is a polynomial of at most the same degree as $q$ (and hence $p$) and $q_n$ converges to $q$ uniformly on $[0,1]$. Since the degree is bounded, it follows that the coefficients of $Q_n$ converge to the coefficients of $q$. In particular, $\deg q_n=\deg q$ for all large enough $n$. Let $a$ be the leading coefficient of $q$ and $a_n$ the leading coefficient of $q_n$. Then $a_n\to a$ and hence the polynomial $$ r_n(x):=\frac a{a_n} q_n(x) $$ converges to $q(x)$ uniformly on $[0,1]$. Choose $n$ so large that $|r_n-q|<c$ on $[0,1]$, then $r_n<p$ on $[0,1]$. Since $r_n$ is already represented in the desired form, it remains to represent $p-r_n$. Observe that $\deg(p-r_n)<\deg p$ and proceed by induction.

Appendix: Why $\deg q_n\le\deg q$.

Since $q_n$ is linear in $q$, it suffices to consider the case $q(x)=x^k$. Let $$ F_0(u,v) = (u+v)^n = \sum_{i=0}^n \binom ni u^i v^{n-i} $$ and $$ F_m = u\cdot \frac{\partial}{\partial u} F_{m-1}, \qquad m=1,2,\dots $$ Then $$ F_k(u,v) = \sum_{i=0}^n i^k \binom in u^iv^{n-i}, $$ hence $q_n(x) = n^{-n} F_k(x,1-x)$. By induction in $k$ one sees that $F_k(u,v)$ is a linear combination of terms of the form $u^j(u+v)^{n-j}$ with $j\le k$. Hence $F_k(x,1-x)$ has degree at most $k$, and so does $q_n(x)$.

Remark. Analyzing the coefficients of the above linear combination, one can see that, as $n\to\infty$ and $k$ fixed, the coefficient at $u^j(u+v)^{n-j}$ goes to 1 and all other ones go to 0. This gives an elementary proof of convergence $q_n\to q$ (coefficient-wise) avoiding the general theory.