 Solution

The assertion is true. The assumed finite multiplicity means that $f_0-w_1$ is not identically zero and has an isolated zero of order $k$ at $z_1$. Choose $\rho >0$ so that $\overline{D(z_1,\rho )}\subset D(0,r)$ and $f_0-w_1$ has no other zeros in this closed disc. On its boundary,The joint holomorphy established above gives $f_{\alpha }\to f_0$ uniformly on that circle as $\alpha \to 0$. For sufficiently small $|\alpha |$, therefore, $|f_{\alpha }-f_0|<\eta$. By Rouché's theorem, $f_{\alpha }-w_1$ has exactly $k$ zeros in $D(z_1,\rho )$, counted with multiplicity. In particular it has at least one zero in $D(0,r)$. This preserves the total local zero count, without asserting that the zeros remain together.