Concerning $C^0[0,1]$ and the $L^1$-Norm. Consider the well known Euler sequence of functions $x^n$ ($n=1,2,3\ldots$) on $[0,1]$. It is clear that it converges against $\chi_{1}$, the characteristic function of the singleton $1$, in the $L^1$-Norm, and that this function does not lie in $C^0[0,1]$ is supposed to show that $C^0[0,1]$ is not complete. However it converges also against $0\in C^0[0,1]$. So is the argument really valid? Problem Fréchet?
EDIT: The main question is: does $x^n\rightarrow\chi_{1}$ in $L^1$ refute that $C^0[0,1]$ is complete? After all we also have $x^n\rightarrow 0\in C^0[0,1]$ in $L^1$?
 A: No, the argument is not valid. Let me give two examples:

Let $X$ be a metric space. Let $Y\subseteq X$ be a subset. Then if $(y_n)$ is a Cauchy sequence in $Y$ such that $y_n \to x_0 \in X\setminus Y$, then $Y$ is not complete. 
This is because for every $y\in Y$, $0 < d(y,x) \leq d(y,y_n) + d(y_n,x)$ by the triangle inequality. So $(y_n)$ cannot converge in $Y$. One can also say that this is a consequence of the standard result that in a metric space, Cauchy sequences can have at most one cluster point. 

Consider the set $Y = [0,1]\subset \mathbb{R}$ and the symbol $a$, and let $X = Y \sqcup \{a\}$. Define the function $d:X\times X \to \mathbb{R}$ by 
$$\begin{cases}
d(y_1,y_2) = |y_1 - y_2| & y_1,y_2 \in Y \\
d(a,a) = 0 \\
d(a,y) = |y| & y\in Y
\end{cases} $$
we have that $d$ is a pseudometric on $X$, and hence a pseudometric on $Y$. It is easy to check that $Y$ is complete with the induced pseudometric. But we have that any converging sequence $y_n \to 0$ also converges toward the point $a$. 

The $L^1$ distance gives a pseudometric (and emphatically not a metric) on the set of measurable functions. Hence as you wrote the argument is invalid for showing that $C^0[0,1]$ is incomplete. You have shown that the Cauchy sequence can converge outside of $C^0[0,1]$, which does not rule out it converging also inside $C^0[0,1]$. And as you have shown, there is a limit in $C^0[0,1]$ for this particular sequence, and so the example is not even salvageable. 

Somewhat related is the following trio of true statements:


*

*If $(X,d)$ is a complete pseudometric space, and $Y\subseteq X$ is closed in the pseudometric topology, then $(Y,d)$ is a complete pseudometric space. 

*If $(X,d)$ is a metric space (not necessarily complete), and $Y\subseteq X$ is such that $(Y,d)$ is a complete metric space, then $Y$ is a closed subset of $X$ in the metric topology. 

*There exists a complete pseudometric space $(X,d)$ with a subset $Y\subseteq X$ such that $(Y,d)$ is a complete pseudometric space and $Y$ is not a closed subset of $X$ in the pseudometric topology. 

A: The main problem is that, while the $L^1$-Norm is in fact a norm on $C^0[0,1]$, it is only a seminorm when considering measureable functions. This is why one usually to considers equivalence classes of summable functions. 
If the initial example doesn't compel you (as the nonunique limit function is a.e. zero and is in the same $L^1$-equivalence class as the zero function), consider the following sequence of functions in $C^0[0,2]$:
$$f_n(x)=\begin{cases}x^n&x\in [0,1]\\ 1 & x\in (1,2]\end{cases}.$$
This sequence of functions converges pointwise and in $L^1$-norm towards $\chi_{[1,2]}$, which is neither continuous nor almost everywhere zero. Even better - there is no contiuous representant in its equivalence class.
A: Let me restate your question to make sure I understand.  You consider the space $C^0[0,1]$ of continuous functions on $[0,1]$ equipped with the $L^1$ norm: 
$$||f|| = \int_0^1 |f(x)|\,dx$$
You are asking: is the sequence $f_n(x) = x^n$ a Cauchy sequence in this space without a limit?
The answer is no.  The sequence is a Cauchy sequence because:
$$||f_n - f_m|| = \left|\frac{1}{n+1} - \frac{1}{m+1}\right|$$
But the sequence converges to $0$:
$$||f_n - 0|| = \frac{1}{n+1} \to 0$$
The characteristic function of $\{1\}$ is not relevant to the structure of $C^0[0,1]$ simply because it is not continuous.  That sort of function becomes a big problem as soon as one makes even a small step away from continuous functions (say, to piecewise continuous functions or measurable functions), but all it really does is prove that $||\cdot||$ is no longer a norm because integration cannot detect removable discontinuities.
A: I think this is a language issue.
You want to make a statement about convergence on the metric space $C[0,1]$ with the metric defined by $d_{L^1}(f,g):=||f-g||_{L^1}$. The concepts of limit and convergence in a metric space are dependant on the set of elements of the space and the metric function defined for those elements. 
$\chi_1$ is not continuous, so it doesn't make sense to speak of it as if it were an element of your metric space of continuous functions. Thus, that $x^n\to \chi_1$ in the metric space $(C[0,1], \, d_{L^1} )$ is a false statement.
Also, that $(C[0,1]\cup\{\chi_1\},\,d_{L^1})$ is a metric space is a false statement   (it doesn't satisfy property 2.). 
What is true is that the sequence of real numbers $\int|x^n-\chi_1|\, dx$ converge to the number $0$ (in the real line with euclidean distance). But this alone is insufficient to assert convergence in $C[0,1]$ with the $L_1$ distance: you need to be talking about distances of elements in your set.
Maybe the confusion arises when you say that the elements converge "in $L_1$". Each time you say that a sequence of elements converge "in $L_1$" you are speaking about a norm and a vector space on which it is defined, even if you would normally omit the latter. The statement "$x^n$ converges to $\chi_1$ in $L^1$" is indeed true when you think the omitted vector space is the space $L^1$ of equivalence classes of measurable functions modulo a.e. equality. Of course, in that space $\chi_1=0$.
