Number of phrases of LZ compression
You don't need to redo the proof for this, simply note that $n$ symbols of an alphabet of size $k$ can be represented with $n \log_2(k)$ bits.  The Lempel-Ziv bound is then:

$\mbox{# phrases} \leq \frac{n log_2 k}{(1-\epsilon_{n \lg k})log_2(n \log_2 k)}$

Dividing numerator and denominator by $\log_2 k$ then gives:

$\mbox{# phrases} \leq \frac{n}{(1-\epsilon_{n \lg k})\left(log_k(n) + \log_2(\log_2(k))/log_2(k)\right)}$

Since $\epsilon_n \longrightarrow 0$ as $n \longrightarrow \infty$, the result follows.