Let $(V, \mathcal{T})$ be a topological vector space. A subset $B\subseteq V$ is bounded if for all opens $O$ containing $0$, there is $t > 0$ such that $B \subseteq tO$.
Next, consider the topological vector space $V = C([0,1], \mathbb{R})$ with topology $\mathcal{T}$ the initial topology generated by $$\{ev_x: C[0,1]\to \mathbb{R}: f \mapsto f(x)\mid x \in [0,1]\}$$
I'm trying to show that if $x \in [0,1]$ and $a < 0 < b$, then the open
$$U:= \{f\in C[0,1]: a < f(x) <b\}$$
is not bounded, but I was unsuccesful. I tried to argue by contradiction, but got nowhere. Then I tried to prove that there is an open $O$ such that for all $t> 0$ we have
$$U \not\subseteq tO$$
or equivalently
$$1/t U = \{f \in C[0,1]: a/t <f(x) < b/t\}\subseteq O$$ but I struggle to find the open $O$ that makes the argument work.
Note that the your topology has a sub-basis of the form $S_{y,c,d}=\{f|ev_y(f)\in (c,d)\}$. So let $y\in [0,1]$, we wish to argue that $U\not\subseteq tS_{y,-1,1}$ for any $t$.
Indeed, assume without loss of generality that $x<y$ and define $g_t$ to be the function $$ g_t(z)=\begin{cases} 0 & z\in [0,x] \\ 2t \frac{z-x}{y-x} & z\in[x,y] \\ 2t & z\in [y,1]\end{cases} $$ Then, $g_t$ is clearly continuous for every $t$ and $g_t\in U$, however, $g_t\not \in t S_{y,-1,1}=S_{y,-t,t}$. Thus, $U$ is not bounded.