Convergence of the sequence $f_{1}\left(x\right)=\sqrt{x} $ and  $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ 
Let $\left\{ f_{n}\right\} $ denote the set of functions on
  $[0,\infty) $ given by $f_{1}\left(x\right)=\sqrt{x} $ and
  $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ for $n\ge1 $.
  Prove that this sequence is convergent and find the limit function.

We can easily show that this sequence is is nondecreasing. Originally, I was trying to apply the fact that “every bounded monotonic sequence must converge” but then it hit me this is true for $\mathbb{R}^{n} $. Does this fact still apply on $C[0,\infty) $, the set of continuous functions on $[0,\infty) $. If yes, what bound would we use?

In a sense, the answer is "yes". You can apply the Monotone Convergence Theorem pointwise.  That is, for a fixed value of $x$, show that the sequence of nonnegative numbers $\bigl(f_n(x)\bigr)$ is bounded above and increasing. Then of course for each $x$, the sequence $\bigl(f_n(x)\bigr)$ converges to some $f(x)$. (You can find $f(x)$ explicitly by taking the limits of both sides of your defining relation for $f_n$.)
Towards showing that $\bigl(f_n(x)\bigr)$ is indeed bounded above, try using the bound $1+2\sqrt x$. 
Perhaps better, as suggested by Did, would be to formally find the limit fuction first, and then show that this serves as an upper bound.
