Without invoking Cesaro-Stolz, "epsilon-delta" argument:
Suppose $z_n \to a$. Let $S_n = \sum_{k=0}^n a_k$. We have
$$
\left|w_n - a\right|=\left|\frac{\sum_{k=0}^n a_k \left|z_k - a\right|}{\sum_{k=0}^n a_k}\right|\leq \frac{1}{S_n}\sum_{k=0}^n a_k \left|z_k - a\right|
$$
Fix $\varepsilon > 0$. Since $z_k\to a$, there is $N$ such that for $n>N$ we have $\left|z_n - a\right|<\varepsilon/2$. Now,
$$
\left|w_n-a\right|\leq \frac{1}{S_n}\sum_{k=0}^N a_k \left|z_k - a\right|+
\frac{1}{S_n}\sum_{k=N+1}^n a_k \left|z_k - a\right|
$$
But $S_n\to\infty$ by assumption (positive terms, does not converge), so the first part tends to $0$ (sum remains fixed). And
$$
\frac{1}{S_n}\sum_{k=N+1}^n a_k \left|z_k - a\right|<
\frac{1}{S_n}\sum_{k=N+1}^n a_k \frac{\varepsilon}{2}=
\frac{\varepsilon}{2S_n}\sum_{k=N+1}^n a_k
$$
Now,
$$
\sum_{k=N+1}^n a_k = \sum_{k=N+1}^n a_k + \sum_{k=0}^N a_k - \sum_{k=0}^N a_k =
\sum_{k=0}^n a_k - \sum_{k=0}^N a_k = S_n - S_N
$$
and so
$$
\frac{1}{S_n}\sum_{k=N+1}^n a_k=\frac{S_n - S_N}{S_n}=1-\frac{S_N}{S_n}<1
$$
therefore
$$
\frac{\varepsilon}{2S_n}\sum_{k=N+1}^n a_k<\frac{\varepsilon}{2}
$$
Hence, for large enough $n$, $\left|w_n-a\right|<\varepsilon$.
Suppose now $z_n\to+\infty$. Fix $M > 0$. From some point, $z_n > M + 1$. Similar argument shows, that part of sum before that point can be neglected (bounded in absolute value by $\varepsilon$), and after that it's big enough to guarantee that limit is greater than $M$:
$$
\frac{1}{S_n}\left|\sum_{k=0}^N a_k z_k\right|\to 0
$$
as $n\to \infty$, since $N$ remains fixed, and $S_n\to\infty$. For large enough $n$ it is thus bounded by $\varepsilon$.
$$
\frac{1}{S_n}\sum_{k=N+1}^n a_k z_k \geq \frac{1}{S_n}\sum_{k=N+1}^n a_k (M+1)=\frac{S_n - S_N}{S_n}(M+1)>M+\varepsilon
$$
eventually, since $S_N/S_n\to 0$. Finally,
$$
w_n=\frac{1}{S_n}\sum_{k=0}^n a_k z_k\geq
\frac{1}{S_n}\sum_{k=N+1}^n a_k z_k-\frac{1}{S_n}\left|\sum_{k=0}^N a_k z_k\right|>M+\varepsilon-\varepsilon=M
$$
Therefore, $w_n > M$ for large enough $n$.
