To prove E[Y] = E[E[Y∣X]] if X is continuous and Y is discrete

Question

Given Y is discrete, E[Y] = $\sum_{y}^{} y * P(Y=y)$ and given X is continuous, E[X] = $\int_{x}^{} x * P(X=x)$. Then, $$E[E[Y|X]] = E[\sum_{y}^{} y * P(Y=y|X)]$$ $$=\int_{x}^{} \sum_{y}^{} y * P(Y=y|X) * f_X(x) dx $$ $$=\sum_{y}^{} y * \int_{x}^{} P(Y=y|X) * f_X(x) dx $$ $$=\sum_{y}^{} y* P(Y=y)$$ $$= E[Y] $$

Is this correct?