How to prove two topologies $\mathcal{T}_1,\mathcal{T}_2$ are not equal Let $C[0; 1]$ be the set of all continuous real-valued functions on $[0; 1]$.
(i) Show that the collection $M$, where $M = \{M(f,\varepsilon ) : \text{$f\in C\left[0; 1\right ]$ and  $\varepsilon $ is a
positive real number}\}$ and $M(f,\varepsilon) =\{g : \text{$g\in C\left[0; 1\right ]$ and
$\int_{0}^{1}\left|f-g\right| < \varepsilon $}\}$, is a
basis for a topology  $\mathcal{T}_{1}$ on $C[0; 1]$.
(ii) Show that the collection $U$, where $U = \{U(f,\varepsilon ) : \text{$f\in C\left[0; 1\right ]$ and  $\varepsilon $ is a
positive real number}\}$ and $U(f,\varepsilon ) =\{g : \text{$g\in C\left[0; 1\right ]$ and
$\sup_{x\in \left[0,1\right]}$$\left|f-g\right|<\varepsilon $}\}$, is a
basis for a topology  $\mathcal{T}_{2}$ on $C[0; 1]$.
(iii) Prove that  $\mathcal{T}_{1}\neq \mathcal{T}_{2}$.
(i)and (ii) are similar by using the property of absolute value $\left|f-g\right|\leq\left|f\right|+\left|g\right|$ for (i) let $M_{1}$ and $M_{2}\in M$ where $M_{1}(f_{1},\varepsilon) =\{g : \text{$g\in C\left[0; 1\right ]$ and
$\int_{0}^{1}\left|f_{1}-g\right|<\varepsilon $}\}$, $M_{2}(f_{2},\varepsilon) =\{g : \text{$g\in C\left[0; 1\right ]$ and
$\int_{0}^{1}\left|f_{2}-g\right|<\varepsilon $}\}$
 then $M_{1}\cap M_{2}=M(\dfrac{f_{1}+f_{2}}{2},\varepsilon )$ so $M$ is a base for $C[0; 1]$.
But I am not sure for (iii) by using the mean value theorem of integrals if $g$ is in some $m\in M$ then there may has no $u\in U$ since $\int_{0}^{1}\left|f-g\right|=\left|(f-g)\right| \left|(\xi)\right|<\varepsilon$ ($\xi \in [0,1]$) but if $\left|(f-g)\right| \left|(\xi)\right|<\sup_{x\in \left[0,1\right]}\left|f-g\right|$ so $g$ is not in some $u$ in $U$. I have no ideal about what to do next.
 A: Parts (i) and (ii) are routine, as we are asked to show the families $M$ and $U$ are open bases for topologies, hereafter called $\mathcal{T}_1$ and $\mathcal{T}_2$.  What needs to be shown for each family is that (1) it covers the space $C[0,1]$, continuous real functions on the unit interval, and (2) that for any "point" $f$ in the intersection $S_1 \cap S_2$ of two members of such a family, there exists a member $S_3$ of that family s.t. $f \in S_3 \subseteq S_1 \cap S_2$.
The covering property is evident for both families from the definitions of the $M$ and $U$ families.  The intersection property is only slightly more work, as we illustrate for the family $M$ of sets $M(f,\varepsilon)$, using triangle inequality arguments.
Suppose $f \in M(f_1,\varepsilon_1) \cap M(f_2,\varepsilon_2)$. 
Since $\int_0^1 |f_1 - f| dx \lt \varepsilon_1$ and $\int_0^1 |f_2 - f| dx \lt \varepsilon_2$, we pick $\varepsilon_3 > 0$ to be the minimum of $\varepsilon_1 - \int_0^1 |f_1 - f| dx $ and $\varepsilon_2 - \int_0^1 |f_2 - f| dx $.  Then for any $g \in M(f,\varepsilon_3)$ we would have (from a triangle inequality) respectively:
$$ \int_0^1 |f_1 - g| dx \le \int_0^1 |f_1 - f| dx + \int_0^1 |f - g| dx \lt \varepsilon_1 $$
$$ \int_0^1 |f_2 - g| dx \le \int_0^1 |f_2 - f| dx + \int_0^1 |f - g| dx \lt \varepsilon_2 $$
and the interesection property is established: $f \in M(f,\varepsilon_3) \subseteq M(f_1,\varepsilon_1) \cap M(f_2,\varepsilon_2)$.
The argument for family $U$ is similar except that the sup-norm rather than the $L^1$-norm is involved.
The more difficult part of the exercise seems to be showing the resulting topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ are different.  The image linked by @Arthur gives the idea; a function can be close to the origin in the $L^1$-norm, but arbitrarily far away in the sup-norm.
It suffices, per comments under the Question, to show that $U(0,1/2)$, an open set in $\mathcal{T}_2$, is not open in $\mathcal{T}_1$.  In particular there does not exist any $\varepsilon \gt 0$ such that $0 \in M(0,\varepsilon) \subseteq U(0,1/2)$, which would be required for $U(0,1/2)$ to be an open neighborhood of the origin in $\mathcal{T}_1$.
To prove this, consider the elementary sequence of real functions $f_n(x) = x^n$ on the unit interval, $n = 1,2,3,\ldots$:
$$ \int_0^1 |0 - x^n| dx = 1/n $$
Since all but finitely many $f_n$ belong to $M(0,\varepsilon)$ but none of them belong to $U(0,1/2)$ (because each *sup*$|f_n(x)| = 1$ on $[0,1]$), clearly $M(0,\varepsilon) \not \subseteq U(0,1/2)$.  Open sets in $\mathcal{T}_2$ are not necessarily open in $\mathcal{T}_1$, so these topologies are not the same.
As it happens, the converse holds however, that open sets in $\mathcal{T}_1$ are always open in $\mathcal{T}_2$. In a case like this we say $\mathcal{T}_2$ is strictly finer than $\mathcal{T}_1$.
