The metric and Kronecker's delta

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I am reading some lecture notes for GR and it is currently showing how we are going to derive the field equations using a metric for a massive free particle with a metric

$$g_{00}=1+\frac{2\phi}{c^2}, \qquad \phi=-\frac{GM}{r} $$

We noted several things

$$ \nabla_{u} ({R^u}_{v} - \frac{1}{2}{\delta^u}_{v}R) =0, \\ \partial_{u}{T^u}_{v} =0 $$

From this we deduced, $$ {R^u}_{v} - \frac{1}{2}{\delta^u}_{v}R = \alpha {T^u}_{v},$$

and therefore,

$$ R_{uv} - \frac{1}{2}g_{uv}R = \alpha T_{uv},$$

However I have a problem with the last line, why has the delta function disappeared without affecting the metric, should the metric not be $g_{uu}$?

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Start from $$R_{\,\,v}^{u} - \frac 12 \delta_{\,\,v}^{u} R = \alpha T_{\,\,v}^{u}$$ multiply both sides by $g_{wu}$ to lower the inedx

$$g_{wu}(R_{\,\,v}^{u} - \frac 12 \delta_{\,\,v}^{u} R) = \alpha g_{wu} T_{\,\,v}^{u}$$ or $$R_{wv} - \frac 12 g_{wv} R = \alpha g_{wu} T_{wv}$$ It remains to ralabel $w\to u$ to get the last line.