To complete Narutaka OZAWA's answer in comment by a concrete example as asked in the OP, here is a bounded linear functional on $c_0$ not attaining its norm w.r.to (an equivalent) strictly convex norm.

On the space $c_0 $ consider a norm $\Vert x\Vert :=\Vert x\Vert _\infty +\Vert x\Vert _2$, which is obtained adding a (weighted) pre-Hilbert norm, hence strictly convex

$$\Vert x\Vert _2:=\sqrt{\sum_{k=1}^\infty 2^{-k}x_k^2}$$
to the standard norm $$\Vert x\Vert_\infty:=\max_{k\ge1}|x_k|.$$ Thus $ \Vert x\Vert _\infty\le \Vert x\Vert\le 2 \Vert x\Vert _\infty$, so $\|\cdot\|$ is equivalent to $\|\cdot\|_\infty$, and it is strictly convex.

Consider the linear form $f\in(c_0)^*=\ell_1$ defined by $f_k:=2^{-k}$. Then for any $x\in c_0\setminus\{0\}$
$$\langle f,x\rangle=\sum_{k=1}^\infty 2^{-k}x_k\le\sqrt{\sum_{k=1}^\infty 2^{-k}}\sqrt{\sum_{k=1}^\infty 2^{-k}x_k^2}=\|x\|_2 $$
and also
$$\langle f,x\rangle<\Big( \sum_{k=1}^\infty 2^{-k}\Big)\|x\|_\infty=\|x\|_\infty =\|x\|-\|x\|_2.$$
Therefore $\displaystyle\langle f,x\rangle<\frac12\|x\|$ for all $x\neq 0$. On the other hand, for the sequence $x^n:=\sum_{k=1}^ne_k=({1,\dots,1},0,0\dots)\in c_0$ one has
$$\frac{\langle f,x^n\rangle}{\|x^n\|}=\frac{ \sum_{k=1}^n 2^{-k}}{1+\sqrt{\sum_{k=1}^n 2^{-k}}}=\frac{ 1- 2^{-n}}{1+\sqrt{1- 2^{-n}}}=\frac12+o(1).$$
So $f$ has dual norm $\displaystyle \frac12$, but it is not attained on the closed unit ball.

And the affine closed hyperplane $C:=\{f=\frac12\}$ has no element of minimum norm (minumum distance from $0$)