Show that $\mathbb{Q}(\sqrt{2},\sqrt{3},\dots,\sqrt{p},\dots)$ is an algebraic extension of $\mathbb{Q}$, for $p$ prime. I have shown that $[\mathbb{Q}(\sqrt{2},\dots,\sqrt{p},\dots):\mathbb{Q}]=\infty$, by showing that 
$$
\mathbb{Q} \subset \mathbb{Q}(\sqrt{2})\subset \mathbb{Q}(\sqrt{2},\sqrt{3})\ldots$$
 is an ascending chain of algebraic extensions of $\mathbb{Q}$. 
 A: Suppose that $\mathbb Q(\sqrt{2},\sqrt{3},\dots,\sqrt{p},\dots)|\mathbb Q$ is not an algebraic extension of $\mathbb Q$. Then there is $\alpha \in \mathbb Q(\sqrt{2},\sqrt{3},\dots,\sqrt{p},\dots)$ such $[\mathbb Q(\alpha):\mathbb Q]=\infty$. But $\alpha = \sum_{i=0}^n q_i\sqrt{a_i}$, where $a_i \in \mathbb{N}$ (and they are square free). The most important thing about this sum is that it is finite. It implies that the set of all different prime numbers which are present in decompositions of all $a_i$ is finite (suppose, that $P$ is the largest among them). Then $\alpha \in \mathbb Q(\sqrt{2},\sqrt{3},\dots,\sqrt{P})$ and it is an contradiction to our assumption.
A: More generally, if $\alpha_1,\alpha_2,\dots,\alpha_n,\dotsc$ is a sequence of algebraic elements over the field $F$, then
$$
K=F(\alpha_1,\alpha_2,\dots,\alpha_n,\dotsc)=
\bigcup_{n\ge 1}F(\alpha_1,\alpha_2,\dots,\alpha_n)
$$
is algebraic over $F$.
Indeed, if $\beta\in K$, then $\beta\in F(\alpha_1,\alpha_2,\dots,\alpha_n)$ for some $n\ge1$. By the dimension formula, $F(\alpha_1,\alpha_2,\dots,\alpha_n)$ is finite dimensional over $F$, so it is algebraic over $F$ and, in particular, $\beta$ is algebraic over $F$.
In your case, $\sqrt{p}$ is algebraic over $\mathbb{Q}$, for any prime $p$.
A: Another approach: Your fields are all inside $\Bbb C$, an algebraically closed field. The subset of all elements of $\Bbb C$ that are algebraic over $\Bbb Q$ is a subfield, since two numbers algebraic over $\Bbb Q$ generate a finite (and thus algebraic) extension of $\Bbb Q$. So this set of all algebraic numbers is a field, obviously containing your infinite extension.
The truth is stronger than that. In fact any compositum of algebraic extensions of a field $F$ is algebraic over $F$. My argument above has made strong use of the existence of an algebraically closed field larger than all the given fields. If you don’t have at your disposal the theorem that every field has an algebraic closure, you’ll have to fall back on some argument much like that of @egreg, but jazzed up with transfinite induction. How do you know that $\Bbb C(t, \{\sqrt{t-a}\}_{a\in\Bbb C})$ is algebraic over $\Bbb C(t)$? You’ve adjoined uncountably many independent square roots.
