Using Banach-Alaoglu theorem on $L^1$

Let $$h_n=n\chi_{[0,1/n]} $$ and $f=1$, $C=\left\{h_n\right\}$.

Then $C$ is bounded and closed in $L^1(0,1)$ (since $\|h_n\|_{L^1}=1$ for all $n$), and for any subsequence $\left\{h_{n_k}\right\}$ we have, for each $t\in (0,1]$, $$\int_0^t h_{n_k}f=\int_0^tn_k\chi_{[0,1/n_k]}\to1$$ Now suppose by contradiction that there exists $h\in L^1(0,1)$ with $$\int_0^tf(s)h(s)\,ds=\int_0^th(s)\,ds=1\qquad \forall t\in (0,1] $$ By absolute continuity of the Lebesgue integral, $$\lim_{t\to 0}\int_0^t h(s)\,ds =0 $$ which is a contradiction.

Notice that if we do not require the equality to hold for all $t \in [0,1]$, but just for e.g. $t=1$, then this counterexample no longer works, as $h=h_n\in C$ (for any fixed $n$) is an admissible limit for any subsequence. I suspect that this weaker statement might actually be true, although it does not seem easy to prove.

Also, it is not hard to check that the same contradiction is obtained if we replace $f=1$ with any $f\in L^{\infty}$ such that $$\lim_{k\to +\infty}n_k\int_0^{1/n_k}f\neq 0 $$ (e.g. any continuous function with $f(0)\neq 0$).

As Lorenzo has pointed out in his fantastic answer this statement may be true if we just want the integral convergence to hold on the fixed interval $[0,1]$. What follows is a partial proof of this statement, in the hope that someone can take it over the line, because I find this to be quite an interesting question.

Let $K$ denote the uniform bound of the sequence in $L^1(0,1)$. Note that by Holder's inequality, for any $f\in L^\infty(0,1)$ and $n\in \mathbb N$, we have $$\left |\int_0^1h_nfdx\right|\leq K\|f\|_\infty.$$ Thus $\left(\int_0^1h_nfdx\right)$ is a bounded sequence in $\mathbb R$ (or $\mathbb C$), so by Bolzano-Weierstrass there exists a convergent subsequence $\left(\int_0^1h_{n_k}fdx\right)$ converging to some $a\in \mathbb R$, with $|a|\leq K\|f\|_\infty$.

What remains to prove is that there exists some $h\in C$ such that $$a=\int_0^1hfdx.$$

We note that $L^\infty(0,1)\subset L^1(0,1)$ and $L^\infty(0,1)$ separates the points of $L^1(0,1)$. The Riesz representation theorem thus implies that there exists some $h\in L^\infty(0,1)$ such that $$\int_0^1hfdx=a.$$ Unfortunately this method does not guarantee that $h\in C$. I believe, if it is possible at all, some rather subtle tools will be needed to show we can find some $h\in C$.