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The acidity of a compound is determined by the stability of its conjugate base and the major structure of enolate has a negative charge on the Oxygen atom. Moreover, the negative charge on the O-atom of ethoxide is increased by the +I effect of $\ce{-CH2CH3}$.
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So, how is ethoxide more stable than enolate?
Taking the data from this answer by ron, the equilibrium constant for the keto-enol tautomerism seen in acetone is
$$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$
$$ \begin{array}{lc} \hline \text{compound} & K_\mathrm{eq} \\ \hline \text{acetaldehyde} & 6 \times 10^{-7} \\ \text{acetone} & 5 \times 10^{-9} \\ \hline \end{array} $$
This means that acetone exists predominantly (if not wholly) in its keto form. For every one molecule of the enol form observed, we can expect $\pu{2E8}$ molecules of the keto form.
Because of this, any deprotonation that takes place in acetone will have to be via breaking a $\ce{C-H}$ bond, and that is not a simple thing to do when you compare it to deprotonating an $\ce{O-H}$ bond in ethanol. Thus ethanol is more acidic than acetone.
