Lower bound on prefix code lengths
I'll try to show this, hope I interpreted the question correctly.

Let $A_k=\{0,1\}^k.$ If $f(n)\leq n+\log n$ for $n$ large enough this implies $$\max\{\ell(c(x)):x \in A_1 \cup A_2\cup \ldots \cup A_k \}\leq k+ \log k,$$ for $k\geq N$, for some finite $N$, where $\ell(c(x))$ is the length of the codeword $c(x)$ assigned to $x.$ So this means that
$$
\sum_{N\leq k ~~} \sum_{x \in A_1\cup A_2 \cup \ldots \cup A_k} 2^{-\ell(c(x))}\geq \sum_{N \leq k~~} \sum_{x \in A_1\cup A_2 \cup \ldots \cup A_k} 2^{-(k+\log k)}
$$
which is equal to
$$
\sum_{N\leq k} |A_1 \cup A_2 \cup \ldots \cup A_k|~ 2^{-(k+\log k)}=
\sum_{N\leq k} (2^{k+1}-1)~ 2^{-(k+\log k)}
$$
which is equal to
$$
\sum_{N\leq k} 2^{1-\log k} - 2^{-(k+\log k)}=2 \sum_{N\leq k} \frac{1}{k} - \sum_{N\leq k}\frac{2^{-k}}{k}=2 \left(\sum_{N\leq k} \frac{1}{k}\right) -\ln 2
$$
which is unbounded, since the harmonic series diverges.