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Is there a way to know the number of real roots, or to have some bound at least to the number of real roots of a continuous function $f(x)$.

I think, intuitively, that the number of real solutions to $f(x)=0$ should be bounded by the number of roots of $f'(x)$ plus one.

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    For a counter-intuitive example use $f(x)=x\sin(1/x)$. A continuous function without bound on the number of roots. $f(x)=x^2\sin(1/x)$ is even differentiable. – 2017-01-02
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    This doesn't work because it is not differentiable at $0$? @LutzL – 2017-01-02
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    You were asking about general continuous functions, no differentiability assumed. And the second function has derivative $0$ at $0$, which is a text-book example for a non-continuous derivative. – 2017-01-02

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Assuming you mean roots over $\mathbb{R}$, the number of roots of $f(X)$ is bounded above by the number of roots $f^\prime(x)$ plus one. This can be seen by using Rolle's theorem. (Assuming $f$ is differentiable)

That is suppose, that $f(x)$ has $k$ roots $x_1,\cdots, x_k$. By Rolles theorem, we know that $f^\prime(x)$ has at least one root between $x_i$ and $x_{i+1}$ for each $i=1,\cdots, k-1$. Hence $f^\prime(x)$ has at least $k-1$ roots and so the statement follows.

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If $f$ is continuously differentiable you are correct. This is a consequence of Rolle's theorem If $f(x)$ takes the same value at two different values of $x$, there is a stationary point between them, so between any two roots there is a root of $f'$. You need to make sure your function satisfies the hypotheses of the theorem. There is no lower bound, though. $f(x)=2+\sin x$ has no roots, but its derivative has infinitely many.