= Solution
The proposed <norm> is finite because every <continuous function> on $[0,1]$ is bounded and integrable. It is nonnegative. If $\|f\|=0$ and $f(x_0)\ne0$ at some point, <continuity> gives a nontrivial interval within $[0,1]$ on which $|f|$ is bounded below by a positive constant, contradicting the zero integral. Thus $\|f\|=0$ implies $f=0$. The converse is immediate. Absolute homogeneity follows from $|cf(x)|=|c||f(x)|$, and the pointwise inequality $|f(x)+g(x)|\leq|f(x)|+|g(x)|$ integrates to the <triangle inequality>. These establish \b[all the norm axioms], so $V$ is a <normed vector space>.
To test the <Cauchy sequence> condition, compare the terms with indices $n$ and $2n$. Put $d_n(x)=\sin(nx)-\sin(2nx)$. The identities $2\sin(nx)\sin(2nx)=\cos(nx)-\cos(3nx)$ and $2\sin^2(nx)=1-\cos(2nx)$ give
$$
\begin{aligned}
\int_0^1d_n(x)^2\,dx
&=\int_0^1\sin^2(nx)\,dx+\int_0^1\sin^2(2nx)\,dx-2\int_0^1\sin(nx)\sin(2nx)\,dx\\
&=1-\frac{\sin(2n)}{4n}-\frac{\sin(4n)}{8n}-\frac{\sin n}{n}+\frac{\sin(3n)}{3n}\longrightarrow1.
\end{aligned}
$$
Because $|d_n(x)|\leq2$, we have $d_n(x)^2\leq2|d_n(x)|$. Therefore
$$
\liminf_{n\to\infty}\|f_n-f_{2n}\|\geq\frac12.
$$
In particular, for all sufficiently large $n$, these arbitrarily late pairs have <norm> distance greater than $1/4$. This proves the <sine sequence is not Cauchy in the integral norm>. Thus \b[the sequence is not Cauchy and does not converge to an element of $V$], since every <norm convergent> sequence is <Cauchy>. This direct separation argument does not rely on <completeness> of $V$.
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