Part (a) of Exercise 13 of first chapter of Rudin's book "Functional Analysis"

What's condition of UBP and what's the conclusion? You need a Banach space, a normed linear space and a linear map between them. Your conclusion is also about the norm of a bounded linear map. Where is the linear map in your question?

Even if you define those $Q_x$, they are linear functionals in the dual space of $C[0,1]$ with the $\|\cdot\|_\infty$ norm. Nothing about these $f\in E$ can you obtain from the boundedness of these functionals.

Besides, what's the definition of a bounded set in some topological vector space? It is not equivalent to a bounded set with respect to some metric. Actually the whole space is already bounded with respect to the metric.

If $B_r(0)$ is an open $0$-nbhd in ($C,\sigma)$, you want to show $\exists t>0$ s.t. $E\subseteq t\cdot B_r(0)$. For some $f\in E$, $f=t\cdot\frac{f}{t}.$ Your goal is to prove $\frac{f}{t}\in B_r(0)$. i.e. $$ \int_0^1 \frac {\frac{|f(x)|}{t}} {1+\frac{|f(x)|}{t}} dx = \int_0^1 \frac{|f(x)|}{t+|f(x)|}dx<r. $$ That is to find some $t$ big enough such that $\int_0^1\frac{|f(x)|}{t+|f(x)|}dx$ can be arbitrarily small for pointwise bounded family of functions. When $t\to\infty$, $\frac{|f(x)|}{t+|f(x)|}\to0$ pointwise and $\frac{|f(x)|}{t+|f(x)|}$ is always bounded by an integrable function. What does this imply? Can you prove that $$ \lim_{t\to\infty}\int_0^1 \frac{|f(x)|}{t+|f(x)|}dx=0? $$ How to make that happen for all $f\in E$ so that you can have the $t$ big enough uniformly? i.e. the following hidden hint.

The pointwise supreme of any family of lower semi-continuous functions is also lower semi-continuous thus measurable.

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Define $\phi(x)=\sup_{f\in E}|f(x)|$. They are all bounded thus $\phi$ is a well defined real function and $\phi$ is lower semi-continuous. Thus $$d(f/t,0)=\int_0^1\frac{|f(x)|}{t+|f(x)|}dx\le\int_0^1\frac{\phi(x)}{t+\phi(x)}dx\to 0\ \ (DCT).$$