The Riemann integral over $[a,b]$ of a continuous function $f$ is $$\int\limits_a^bf(x)dx=\lim\limits_{\delta\rightarrow 0} \sum\limits_{i=0}^{n-1} (x_{i+1}-x_i)f(c_i)$$ where $c_i\in[x_i,x_{i+1}]$ and $\delta=\sup(x_{i+1}-x_i)$. The set $[a,b]$ is uncountable. When $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ goes to $1$ and so the number of terms in the sum above is countable. How could a counatble set form a partition of an uncountable set (the sets $]x_i,x_{i+1}[$ form a partition of $]a,b[$).
Now I suspect that there's something wrong in "when $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ goes to $1$" but even if I replaced it with "when $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ goes to an integer $k$" or "when $\delta\rightarrow 0$, the number of elements in a set $[x_i,x_{i+1}]$ is countable" (I don't know why it isn't $1$) I still have the same problem.
The most simple limit sum to think of is $$\sum_{k=1}^n \frac 1n =1$$ as $1/n\to 0^+$. Even being that simple, it will clear up the situation. Staring at it, you might say
Well, in the limit $1/n$ goes to $0$ and there is a countable number of terms in this sum. There is clearly something wrong. How could a countable number of zeros sum up to one?
This question is more fundamental than integrals or Riemann sums, it is a question on the behaviour of limits. The formal definition of a limit will rescue us. To refresh our minds, a quick reminder: we say $\lim\limits_{x\to 0}f(x)=L$ if, and only if, for every $\varepsilon>0$ there is $\Delta >0$ such that $$0<|x|<\Delta\implies |f(x)-L|<\varepsilon.$$
Note that we never mentioned what would happen if $x=0$. It does not concern our definition, and that's the fundamental mistake here. We should not judge the sum in the hypothetical situation $1/n=0$, but only when $0<1/n - 0 <\Delta$ for some finite $\Delta$. The paradox disappears, because at any such interval, $\sum 1/n = 1$.
At any finite $\delta=\sup(x_{i+1}-x_i)$, the cardinality of $[x_i, x_{i+1}]$ is definitely not approaching $1$, even though it is $1$ when $\delta =0$. This illustrates a crucial difference between $\delta\to 0$ and $\delta =0$ that cannot be overestimated, and you will highly benefit from thinking about it. To finish the discussion, you might be interested in rereading some content on continuous functions.