This is the exercise 12 on page 403 of Analysis I of Amann and Escher. I tried to solve this exercise some time ago with no success and I come back again to try to find a solution, however I rediscovered that I dont have a clue about how to solve it.
The exercise says:
Let $E$ a Banach space and $f\in C([0,1],E)$. Show that the sequence $(B_n(f))$ of Bernstein polynomials for $f$, $$B_n(f):=\sum_{k=0}^nf(k/n)B_{n,k},\quad n\in\Bbb N,$$ converges in $C([0,1],E)$ (and hence uniformly in $[0,1]$) to $f$. (Hint: for suitable $\delta>0$ consider $|x-n/k|\ge\delta$ and $|x-n/k|<\delta$, and use the previous exercise).
Some context: the exercise appear in the chapter about Weierstrass approximation theorem (however I dont see a direct relation with the theorem) and prior to any theory about integration. Here $B_{n,k}:=\binom{n}k X^k(1-X)^{n-k}$ is an elementary Bernstein polynomial. The previous exercise, that the hint is talking about, have the following identities
$$\sum_{k=0}^n B_{n,k}=1,\qquad \sum_{k=0}^n kB_{n,k}=nX\\ \sum_{k=0}^n k(k-1)B_{n,k}=n(n-1)X^2,\qquad\sum_{k=0}^n (k-nX)^2B_{n,k}=nX(1-X)$$
My work so far about the exercise: let $f\in C([0,1],E)$ and $B_n(f):=\sum_{k=0}^n f(k/n)B_{n,k}$. Then $f$ is uniformly continuous, so for some chosen $\epsilon>0$ there is a $\delta>0$ such that $$ |x-y|<\delta\implies|f(x)-f(y)|<\epsilon,\quad x,y\in [0,1]\tag1 $$ and we want to show that $\lim_{n\to\infty} B_n(f)=f$. Now observe, using the identities of the previous exercise, that $$ |f(x)-B_n(f)(x)|=\left|f(x)\sum_{k=0}^n B_{n,k}(x)-\sum_{k=0}^n f(k/n)B_{n,k}(x)\right|\\\le\sum_{k=0}^n|f(x)-f(k/n)|B_{n,k}(x)\tag2 $$ Now my idea was to bound $(2)$ is some way such that when $n\to\infty$ then it shows that $B_n(f)(x)\to f(x)$. Its clear from $(1)$ that when $k/n\in\Bbb B(x,\delta)$ then we have some addends on $(2)$ bounded by $\epsilon>0$.
However for the other addends I dont have a clue about how to bound them appropriately. A random idea: if $f$ would be Lipschitz continuous then we can see that
$$|f(x)-f(k/n)|< K|x-k/n|=\frac{K}n|nx-k|\tag3$$
for some $K>0$, and from $(2)$ we will had
$$|f(x)-B_n(f)(x)|\le\sum_{k=0}^n|f(x)-f(k/n)|B_{n,k}(x)\\\le\frac{K}n\sum_{k=0}^n|nx-k|B_{n,k}(x)\le \frac{K}{n}\sqrt{\sum_{k=0}^n\big((nx-k)B_{n,k}(x)\big)^2}\cdot\sqrt{n}\\\le \frac{K}{\sqrt{n}}\sqrt{\sum_{k=0}^n\big(nx-k)^2B_{n,k}(x)}=K\sqrt{x(1-x)}\le K\tag4$$
where I used the Cauchy-Schwarz inequality and the fact that $(B_{n,k}(x))^2\le B_{n,k}(x)$ because $0\le B_{n,k}(x)\le 1$ for all $x\in[0,1]$. However $f$ is not necessarily Lipschitz continuous and $(4)$ is a useless bound.
Some help will be appreciated, thank you.
Hint: For the part that you cannot bound, consider: $$\delta^2\sum_{k=1}^n|f(x)-f(k/n)|B_{n,k}(x)\leq 2M\sum_{k=1}^n|x-\frac kn|^2B_{n.k}(x),$$
where $M$ is the supremum of $f$ over $[0,1].$ Now, those identities you proved earlier will help.
Note that you can extend the proof for only bounded functions $f$ that is not necessarily continuous on all of $[0,1].$ Then, the theorem is true for any continuity point of $f.$