If the operator $T$ is defined by $Tf(x)=\int_0^xf(t)\,dt$, show that $Tf \in C[0,1]$ 
Consider the operator $T$ on $L^2[0,1]$ defined by $Tf(x)=\displaystyle \int_0^xf(t)\,dt.$ Show that $Tf \in C[0,1].$ 

I have one question before this:
What are the implications between $L^p$ spaces, i.e if $f \in L^p$ does this imply that $f \in L^{p+1}$?
My attempt:
Assume $Tf$ is not in $C[0,1]$, so there exists $y \in [0,1]$ and $\epsilon>0$ such that for all $\delta >0$, we can find $x_0$, with $|x-x_0|<\delta$ but 
$$\bigg\rVert\int_0^x f-\int_0^{x_0} f\,\bigg\rVert>\epsilon$$
WLOG assume $x>x_0$, so $$\bigg\lVert \int_{x_0}^x f\,\bigg\rVert \geq \epsilon.$$
So for $\delta_n=\frac{1}{n},$ we can find $x_n \in [0,1]$ such that $|x-x_0|<\frac{1}{n}$ and $$\bigg\lVert\int_{x_n}^xf\,\bigg\rVert \geq \epsilon$$
I don't know if that will lead me to a contradiction.
I would appreciate any help or hints with that.

Your first question is actually relevant here: the relevant inclusion is that a $L^p$ function on a finite measure space is in $L^r$ for all $r<p$. Thus a $L^2([0,1])$ function is also in $L^1([0,1])$. This is the main ingredient you need, along with the theorem that if $f \in L^1$ then for all $\varepsilon > 0$ there exists $\delta > 0$ such that if $\mu(A)<\delta$ then $\int_A |f| d \mu < \varepsilon$. 
One way to prove this theorem more or less "by hand" is to use the bounded convergence theorem: given $g$ such that $\| f - g \|_{L^1} < \varepsilon/2$ and $|g| \leq M=M(\varepsilon)$, you have $\int_A |f| d \mu \leq \int_A |g| d \mu + \int_A |f-g| d \mu < M \mu(A) + \varepsilon/2$, so you can select $\delta=\varepsilon/(2M)$.
Another way to proceed is to use the dominated convergence theorem, by noting that $f(x) 1_{[0,x_n]}(x) \to f(x) 1_{[0,x_0]}(x)$ pointwise if $x_n \to x_0$.
