##\delta_\mu{}^\lambda = 1## if ##\mu = \lambda## and 0 otherwise.
Only repeated indices are contracted, perhaps writing out the summations explicitly makes it clearer:
$$\eta_{\mu\nu} \eta^{\nu\lambda} \equiv \sum_{\nu = 1}^4 \eta_{\mu\nu} \eta^{\nu\lambda}$$
which is equal to 1 if ##\mu = \lambda## and 0 otherwise. In matrix notation, it basically says ##\eta^{-1} \eta = \mathbb{I}##.
If you also contract the two free indices, you get
$$\eta_{\mu\nu} \eta^{\nu\mu} \equiv \sum_{\mu = 1}^4 \sum_{\nu = 1}^4 \eta_{\mu\nu} \eta^{\nu\mu}$$
Replacing the inner sum by the delta, this is
$$\cdots = \sum_{\mu = 1}^4 \delta_\mu{}^\mu$$
and since the upper- and lower index are the same, the delta evaluates to 1 for every value of mu:
$$\cdots = \sum_{\mu = 1}^4 1 = 4$$
As $$\eta_{\mu\nu} \eta^{\nu\lambda}$$ is the ##(\mu, \lambda)## component of the matrix ##\eta^{-1} \eta##, in matrix notation the double sum this basically says
$$\sum_{i = 1}^4 (\eta^{-1} \eta)_{ii} = 4$$
which is indeed the trace of the 4x4 identity matrix.
