Regarding a function's fixed point 
Let $f : C[0,\frac{1}{2}] \to C[0,\frac{1}{2}]$ be a map given by 
  $$ (f(x))(t) = 1 + \int_{0}^{t}x(s)\phantom{.}ds$$
Define $x_{n} = f(x_{n-1})$ starting from $x_{1}(t) = 1 + t$. Find the general form of $x_{n}$ and prove it by induction.

Let $x_{1}(t) = 1 + t$. Then,
$$x_{2}(t) = f(x_{1}) = 1 + \int_{0}^{t} (1 + s)\phantom{.}ds = 1 + t + \frac{1}{2}t^{2}$$
$$x_{3}(t) = f(x_{2}) = 1 + \int_{0}^{t} \bigg(1 + s + \frac{1}{2}s^{2}\bigg)\phantom{.}ds = 1 + t + \frac{1}{2}t^{2} + \frac{1}{6}t^{3}$$
$$x_{4}(t) = f(x_{3}) = 1 + \int_{0}^{t} \bigg(1 + s + \frac{1}{2}s^{2} + \frac{1}{6}s^{3}\bigg)ds = 1 + t + \frac{1}{2}t^{2} + \frac{1}{6}t^{3} + \frac{1}{24}t^{4} $$
Claim: $$ x_{n}(t) = 1 + t + \frac{1}{2}t^{2} + \cdots + \frac{1}{n!}t^{n} = \sum_{i=0}^{n}\frac{1}{i!}t^{i}$$
Base case: for $n=1$, we have $x_{1}(t) = 1 + t$, which we know to be true by assumption. Assume true for some $n = k$. Then, we have
$$
    x_{n+1}(t) = 1 + t + \frac{1}{2}t^{2} + \frac{1}{6}t^{3} + \cdots + \frac{1}{n!}t^{n} + \frac{1}{(n+1)!}t^{n+1}
               = x_{n}(t) + \frac{1}{(n+1)!}t^{n+1}
$$
but I don't think this is finished here. Are there any hints for this induction proof?
Finally, 

Find the unique fixed point of $f$ as a limit of $\{x_{n}\}$.

How may I proceed here?
 A: For induction, don't replace it with $x_n (t)$.  Replace the sequence with $\sum_{i=0}^{n}\frac{1}{i!}t^{i}$
Then $x_{n+1} = \sum_{i=0}^{n}\frac{1}{i!}t^{i} + \frac{1}{(n+1)!}t^{n+1} = \sum_{i=0}^{n+1}\frac{1}{i!}t^{i}$ which is what you're trying to show.
This process is called Picard iteration btw.  
Consider any subsequence of functions $f_j$, $f_k$ on this compact space [0,1/2].  Since for any $\epsilon$>0 there exists $j,k$ s.t. $|f_k - f_k| < \epsilon$, it follows that every subsequence converges.  
You then need to prove that this point is unique.  The best way is to show that the function is Lipschitz continuous, which I will leave to you (OR better yet do what Saibal said below and just show this Cauchy Sequence is also uniformly Cauchy)
A: The part about finding general form of $x_n$ is (EDIT: almost; see Phillip's answer) complete. The fixed point part is left to prove.
First let's find the limit of the function sequence $\left\{x_n\right\}_n$. Fix $t$, then $\left\{x_n(t)\right\}_n$ forms a Cauchy sequence converging to $\exp(t)$. Thus $x_n$ converges to $\exp$ pointwise.
Now, you have
$$x_{n+1}(t) = 1 + \int_{0}^{t} x_n(s)\ ds$$
Taking $n \rightarrow \infty$,
$$\exp(t) = 1 + \lim_{n \rightarrow \infty} \int_{0}^{t} x_n(s)\ ds$$
If you can take the limit inside integration, you will get
$$\exp(t) = 1 + \int_{0}^{t} \exp(s)\ ds = f(\exp)(t),$$
in other words, $\exp$ is a fixed point of $f$. But to take the limit inside, pointwise convergence is not enough, you need uniform convergence. To prove this claim, proceed as follows:
Since $t \in \left[0, \frac{1}{2}\right]$, we have $t^n < 1$ for all $n > 0$. Thus
$$\lvert x_{n+1}(t) - x_n(t) \rvert < \frac{1}{(n + 1)!} \quad\forall\,t$$
i.e. $\left\{x_n\right\}_n$ is uniformly Cauchy, hence convergence is uniform.
