[See here and here for an introduction to the proof. They are explicitly worked special cases]
As you surmised, induction works, employing our prior Lemma (case $\rm\:n = 2\:\!).\:$ Put $\rm\:K = \mathbb Q\:$ in
Theorem $\rm\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} = k\in K\ \Rightarrow \sqrt{c_i}\in K\:$ for all $\rm i,\:$ if $\rm\: 0 < c_i\in K\:$ an ordered field.
Proof $\: $ By induction on $\rm n.$ Clear if $\rm\:n=1.$ It is true for $\rm\:n=2\:$ by said Lemma. Suppose that $\rm\: n>2.$ It suffices to show one of the square-roots is in $\rm K,\:$ since then the sum of all of the others is in $\rm K,\:$ so, by induction, all of the others are in $\rm K$.
Note that $\rm\:\sqrt{c_1}+\cdots+\sqrt{c_{n-1}}\: =\: k\! -\! \sqrt{c_n}\in K(\sqrt{c_n})\:$ so all $\,\rm\sqrt{c_i}\in K(\sqrt{c_n})\:$ by induction.
Therefore $\rm\ \sqrt{c_i} =\: a_i + b_i\sqrt{c_n}\:$ for some $\rm\:a_i,\:\!b_i\in K,\:$ for $\rm\:i=1,\ldots,n\!-\!1$.
Some $\rm\: b_i < 0\:$ $\Rightarrow$ $\rm\: a_i = \sqrt{c_i}-b_i\sqrt{c_n} = \sqrt{c_i}+\!\sqrt{b_i^2 c_n}\in K\:\Rightarrow \sqrt{c_i}\in K\:$ by Lemma $\rm(n=2).$
Else all $\rm b_i \ge 0.\:$ Let $\rm\: b = b_1\!+\cdots+b_{n-1} \ge 0,\:$ and let $\rm\: a = a_1\!+\cdots+a_{n-1}.\:$ Then
$\rm \sqrt{c_1}+\cdots+\!\sqrt{c_{n}}\: =\: a+(b\!+\!1)\:\sqrt{c_n} = k\in K\:\Rightarrow\:\!\sqrt{c_n}= (k\!-\!a)/(b\!+\!1)\in K$
Note $\rm\:b\ge0\:\Rightarrow b\!+\!1\ne 0.\:$ Hence, in either case, one of the square-roots is in $\rm K.\ \ $ QED
Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $\:\sqrt{2} + (-\sqrt{2})\in \mathbb Q\:$ but $\rm\:\sqrt{2}\not\in\mathbb Q.\:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $\,0$.
See also this post on linear independence of square-roots (Besicovic's theorem).