A sequence of functions $\{f_n(x)\}_{n=1}^{\infty} \subseteq C[0,1]$ that is pointwise bounded but not uniformly bounded.

You had the right idea, but don't let the spike have a gentle slope on the right. Try $$ f_n(x) = \cases{n^2 x & if $x < 1/n$\cr n^2 (2/n - x) & if $1/n \le x \le 2/n$\cr 0 & otherwise\cr}$$

Consider the $C^\infty[0,1]$ functions $f_n$ defined by $$ f_n(x)=n^2x^n(1-x). $$ Then $f_n$ is maximum at $x_n=n/(n+1)$ and $f_n(x_n)\sim n/\mathrm e$ hence $\|f_n\|_\infty\to\infty$ when $n\to\infty$, but, for every $x$ in $[0,1]$, $f_n(x)\to0$ when $n\to\infty$ hence the sequence $(f_n(x))_{n\geqslant0}$ is bounded.

First, note that your pointwise bound $\varphi(x)$ must be discontinuous, else it would be bounded on $[0,1]$ and you'd have a uniform bound. Now, consider the function $$f(x)=\begin{cases}\frac 1 x&x\neq 0\\{}\\ 0&x=0\end{cases}$$

We know this is unbounded near $x=0$. At each $1,1/2,\ldots$ this admits a tangent line that is strictly below the function. This tangent is $$T_n(x)=f(n^{-1})+f'(n^{-1})\left(x-\frac 1n\right)\\=2n-n^2x$$

This certainly will follow $f$s behaviour. Then, we continuously join this tangent point to the origin, using $$T'_n(x)=n^2x$$

and to keep $T_n$ positive we cut it off at its root, $2n^{-1}$. Note this gives Robert's solution, and indeed this idea will work for any convex function that goes unbounded near the endpoints.