There's no way this can be literally true:

$$[M,N]^{diff} = [M,N]^{cont}$$

Most of the continuous functions from $M$ to $N$ are not differentiable. So there's no way the above equality can be an equality of sets. I think what you want to ask is if the inclusion:

$$[M,N]^{diff} \to [M,N]^{cont}$$

a bijection? This is answered affimatively in Hirsch's "Differential Topology" textbook. It boils down to a smoothing argument, that every continuous function can be uniformly approximated by a $C^\infty$-smooth function and the smoothing is unique up to a small homotopy. The argument goes further, to state the the space of continuous functions has the same homotopy-type as the space of $C^\infty$ functions. The smoothing argument can be done with bump functions and partitions of unity, and also via a standard convolution with a bump function argument ("smoothing operators").