= Solution
Two <Lipschitz equivalent norms> $N_1,N_2$ satisfy $cN_1(v)\leq N_2(v)\leq CN_1(v)$ for all $v$, for constants $0<c\leq C<\infty$. These inequalities give the same convergent sequences in both norms. A subset of a <metric space> is <closed> exactly when it contains the limits of all convergent sequences from that subset. Therefore \b[closedness is the same in the two norms].
For finite <dimension>, fix a <basis> $e_1,\ldots,e_d$ and let $x=\sum_jx_je_j$. For any norm $N$,
$$
N(x)\leq\sum_j|x_j|N(e_j)\leq C\left(\sum_jx_j^2\right)^{1/2}.
$$
The reverse triangle inequality then makes $N$ continuous in the Euclidean coordinate norm. If no positive lower bound existed on the Euclidean unit sphere, choose $x_m$ there with $N(x_m)\to0$. By the <Bolzano-Weierstrass theorem>, a subsequence converges to $x$ on the unit sphere. Continuity would give $N(x)=0$, although $x\ne0$, a contradiction. Thus $N$ is bounded below by a positive multiple of the Euclidean norm. Comparing any two norms through that coordinate norm proves \b[all norms in finite dimension are Lipschitz equivalent].
On $C[0,1]$, nonnegativity, absolute homogeneity and the triangle inequality for $\|f\|_1=\int_0^1|f|$ follow from the corresponding real-number properties and integration. If a continuous $f$ is nonzero at a point, it has absolute value bounded below by a positive constant on a nontrivial one-sided or two-sided interval around that point. Its integral is then positive. Hence $\|f\|_1=0$ implies $f=0$, completing the <norm> axioms.
For $n\geq2$, put $g_n(x)=\max\{1-n|x-1/2|,0\}$ and $f_n=1-g_n$. Then $f_n(1/2)=0$, but
$$
\|f_n-1\|_1=\int_0^1g_n(x)\,dx=\frac1n\longrightarrow0.
$$
The limit function $1$ is not in the indicated <point evaluation> kernel. Consequently \b[that set is not closed in the integral norm]. This is the <point-evaluation kernel is not closed in the integral norm> phenomenon.
Finally $\|g_n\|_\infty=1$ while $\|g_n\|_1=1/n$. Thus no constant can bound the <uniform norm> by a multiple of the integral norm. Although $\|f\|_1\leq\|f\|_\infty$ on this unit interval, the converse comparison fails: \b[the two norms are not Lipschitz equivalent]. The finite-dimensional theorem cannot be applied to this infinite-dimensional function space.
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