On the other hand, we can also write that $$g_{\mu\nu}\delta^{\nu\alpha} = \delta_{\mu}^{\ \alpha}.$$
This is wrong. The right version directly follows from a plain application of the definition of Kronecker delta, $$g_{\mu\nu}\delta^{\nu\alpha} = g_{\mu\ \alpha}.$$$$g_{\mu\nu}\delta^{\nu}{}_\alpha = g_{\mu\ \alpha}.$$
Edit.
Maybe I got your point. When you write tensor equations using coordinates, you often forget the base you're referring to. Let's analyze the effect of metric tensor on a vector
$$v^{\mu} = g^{\mu\nu} v_{\nu} \ ,$$
that are the $\mu$-th components of the tensor expression
$$\mathbf{v}= v^{\mu} \mathbf{b}_{\mu} = g^{\mu\nu} v_{\nu} \mathbf{b}_{\mu} = \underbrace{g^{\mu\xi} \mathbf{b}_{\mu} \mathbf{b}_{\xi}}_{g \hspace{-3pt} g} \cdot \underbrace{ v_{\nu} \mathbf{b}^{\nu}}_{\mathbf{v}} = g \hspace{-3pt} g \cdot \mathbf{v} \ ,$$
i.e. we've found out that the actual identity tensor is the metric tensor.
Once established that, it's not hard to prove that the identity tensor can be written as well as
$$\mathbb{I} = \delta_{\mu}^{\ \nu} \mathbf{b}^{\mu} \mathbf{b}_{\nu}\ ,$$
where $\delta_{\mu}^{\ \nu}$ are the components of the Kronecker delta, that result to be the components of the identity tensor in mixed base. Using the "law for raising or lower indices", it's easy to prove that this tensor is the metric tensor,
$$\mathbb{I} = \delta_{\mu}^{\ \nu} \underbrace{\mathbf{b}^{\mu}}_{g^{\mu\xi} \mathbf{b}_{\xi}} \mathbf{b}_{\nu} = \underbrace{\delta_{\mu}^{\ \nu} g^{\mu\xi}}_{g^{\nu \xi}} \mathbf{b}_{\xi} \mathbf{b}_{\nu} = g^{\nu \xi} \mathbf{b}_{\xi} \mathbf{b}_{\nu} = g \hspace{-3pt} g \ . $$
It's should be also clear now that $$\delta_{\mu}^{\ \nu} = g^{\nu \xi} g_{\xi \mu} = g_{\mu}^{\ \nu} \ .$$
Edit - details. Some details about reciprocal bases
Reciprocal base. Given a vector base $\{ \mathbf{b}_\mu \}_{\mu=1:N}$, it's (often) possible to define it's reciprocal bases as $\{ \mathbf{b}^{\nu} \}_{\nu=1:N}$, as that set of vectors for which
$$\mathbf{b}^{\nu} \cdot \mathbf{b}_{\mu} = \delta_{\mu}^{\nu}$$
holds. Given a vector (the same holds for tensors of generic rank), it can be written as a linear combination of the vectors of the two bases,
$$\mathbf{v} = v^{\mu} \mathbf{b}_{\mu} = v_{\nu} \mathbf{b}^{\nu}$$
Metric tensor. If you define the metric tensor through the definition of its components,
$$g_{\mu\nu} = \mathbf{b}_{\mu} \cdot \mathbf{b}_{\nu} \qquad , \qquad g^{\mu\nu} = \mathbf{b}^{\mu} \cdot \mathbf{b}^{\nu}$$,
it's easy to prove that its components are what you need to "raise or lower indices".
- Vectors of the bases transform like $$\mathbf{b}^{\nu} = g^{\nu \mu} \mathbf{b}_{\mu} \ .$$ This can be proved 1. taking the dot product with vector $\mathbf{b}_{\xi}$, 2. recalling the definition of reciprocal base and 3. exploiting completeness of the bases $$\mathbf{b}_{\xi} \cdot \mathbf{b}_{\mu} = g_{\nu \mu} \underbrace{\mathbf{b}_{\xi} \cdot \mathbf{b}^{\nu}}_{\delta_{\xi}^{\nu}} \qquad \rightarrow \qquad \mathbf{b}_{\xi} \cdot \mathbf{b}_{\mu} = g_{\xi \mu} \hspace{10pt} \square$$
- components of vectors transform like $$v^{\nu} = g^{\nu \mu} v_{\mu} \ .$$
