Can every irrational number be written in terms of finitely many rational numbers?

A little extra detail, you have included a function, square root. Suppose we add in a finite vocabulary of functions that can also be used, logarithm base $e,$ exponential, trigonometry, inverse trig, hyperbolic trig, your favorite list of less "elementary" functions suh as hypergeometric, Lambert W, whatever.

Any expression is still a finite string combining rational numbers and a fixed alphabet of functions. As a result, the outcome is a countable list of numbers. So, still an uncountable set not accounted for.

The answer is no. The number of expressions which can be formed out of finite combinations of rational numbers is countable, whereas the number of irrational numbers is uncountable, so it cannot be possible.

No, by a counting argument that's worth remembering: There are $\mathfrak{c}=2^{\aleph_0}$ real numbers, but the set of finite sets of rational numbers is countable. The proof on this last statement is relatively straightforward: since we can map rational numbers to whole numbers, we can map finite sets of rationals to finite sets of whole numbers. But now, to each finite set of whole numbers $\{a_1, a_2, a_3, \ldots, a_n\}$ — which we can obviously assume is given in increasing order — associate the number $2^{a_1}3^{a_2}5^{a_3}\ldots p_n^{a_n}$ where $p_n$ is the $n$th prime. You should be able to convince yourself that this mapping is one-to-one, and so there can only be countably many finite sets of rationals.

In fact, this argument shows that 'almost all' reals have no finite description — whether in terms of rationals, roots of algebraic equations, or even an English description like 'the tenth $x\gt 0$ where $e^x\sin(x) = 1.5$' — at all.