By Fermat's sum of sqaure theorem $p=4k+1,\text{ }\exists k\Longleftrightarrow p=a^2+b^2$ $$\sum_{a^2+b^2= n\le x} 1 \ge \sum_{p=a^2+b^2\le x} 1 \gg \dfrac{x}{\log x}$$ and we have $$\sum_{n=a^2\le x}1=\sqrt{x}+O(1)$$ Therefore, for some $0<\delta<1$ $$\sum_{n=a^2+b^2+c^2\le x}1 \ge \left(\sum_{n=a^2+b^2\le x-x^\delta}1\right)\left(\sum_{n=a^2\le x^\delta} 1\right)\gg \dfrac{x}{\log x}\cdot x^{\delta/2}\gg x $$ By Scnhirelman's theorem, we obtain that any natural number can be expressed as a sum of at most $k$ squares, for some constant $k$.
Is it possible to conclude like this?
This argument is not correct. Specifically, the first inequality in the last equation is not correct, because a number can be written as a sum-of-two-squares plus a square in many different ways. A more blatant form of the same error is the claim that the number of integers up to $x$ that can be written as the sum of two integers is $$ \sum_{n=a+b\le x} 1 \ge \bigg( \sum_{a\le x/2} 1 \bigg) \bigg( \sum_{b\le x/2} 1 \bigg) \gg x^2, $$ which is clearly false since the left-hand side is at most $x$. (Indeed, note that your last equation actually proposes to prove that the number of sums-of-three-squares up to $x$ is $\gg x^{1+\delta/2}$ for some $\delta>0$, which is certainly false.)