= Solution
For $f\in C[0,1]$, define its <Bernstein polynomial>
$$
B_nf(x)=\sum_{k=0}^nf\left(\frac kn\right)\binom nkx^k(1-x)^{n-k}.
$$
If $X_{n,x}\sim\operatorname{Bin}(n,x)$, then
$$
B_nf(x)=\mathbb E f(X_{n,x}/n),
\quad
\mathbb E(X_{n,x}/n)=x,
\quad
\operatorname{Var}(X_{n,x}/n)=\frac{x(1-x)}n\leq\frac1{4n}.
$$
Let $M=\|f\|_\infty$. By <uniform continuity>, for any $\varepsilon>0$ choose $\delta>0$ such that $|f(u)-f(v)|<\varepsilon/2$ whenever $|u-v|<\delta$. Splitting the expectation according to that event and using <Chebyshev's inequality> gives, uniformly in $x$,
$$
|B_nf(x)-f(x)|
\leq\frac\varepsilon2+2M\mathbb P(|X_{n,x}/n-x|\geq\delta)
\leq\frac\varepsilon2+\frac{M}{2n\delta^2}.
$$
For sufficiently large $n$ this is below $\varepsilon$. Thus $B_nf\to f$ uniformly on $[0,1]$. An affine change of variable treats every compact interval, proving the <Weierstrass approximation theorem>.
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