Let $f \in C_1([0,1])$
$$\int |f| d\lambda \leq \int \max |f|\, d\lambda, \text{ so }\|f\|_1 \leq \|f\|_2.$$
Now, since $f'$ is continuous on closed interval $[0,1]$ it is bounded. Therefore for each $x \in [0,1]$ we derive
$$
\begin{align}
\|f\|_2 &= \max|f| + \max|f'|\newline
&\leq |f(x)| + \max|f'| + \max|f'|\newline
&\leq |f(x)| + 2\|f\|_1 - 2\int |f|d\lambda\newline
&\leq |f(x)| + 2\|f\|_1.
\end{align}
$$
However we know that $\exists_{x_0} |f(x_0)| \leq \int |f|\, d\lambda$, so we have $|f(x_0)| \leq \|f\|_1$ and finally
$$\|f_2\| \leq 3\|f\|_1.$$
Why $\max|f| \leq |f(x)| + \max |f'|$?
Let $x,y \in [0,1]$. Then $\exists_{c \in [0,1]}$ such that $$\big| |f(x)| - |f(y)|\big| \leq |f(x) - f(y)| \leq |f'(c)| \leq \max |f'|$$ by the mean value theorem.