in order to prove that $$ \lim_{n\to\infty} [a_0,a_1,a_2 \ldots] = \sqrt2 $$ where $a_0=1$ and $a_i=2$ for $ i \ge 1$
I assumed that $ \lim_{n\to\infty} [a_1,a_2 \ldots]$ converges and denoted it as $x$.
this gives out the solution pretty fast using the fact that $x = 2 + \frac{1}{x} $
my question is , is it enough? or do I actually need to prove convergent before I assume it?
I dont want to use the fact that a continued fraction converges if its periodic, I want to know if this argument is valid without knowing convergence
I would say that no, this is generally not enough. As an example, take the sequence defined by:
$$a_0=0\\a_{n+1}=1-a_n$$
This is just a fancy way of saying the sequence is $0,1,0,1,\dots$
For the classical notion of limits, this sequence does not have a limit and does not converge. However, if we were to use your approach and assume it converges to some limit $L$, we'd find that $L=1-L$, implying $L=\frac12$.
There are some different limit definitions which attempt to assign meaning to these values (much like there are different summation techniques which attempt to attach meaning to divergent series). Under these definitions, it is actually true $L=\frac12$ (which is not to say there can't be some other counter-example to your argument, even on that scenario). That said, for classical analysis this is simply flat-out wrong.