The dual space of $c$ is $\ell^1$

The last argument where you say that $f(x)$ would be unbounded does not seem valid since you only have $f(x)\leq \Vert x\Vert_\infty\sum_{k=1}^\infty|f(e_k)|$. If $\sum_{k=1}^\infty |f(e_k)|=\infty$, you don't get any absurd. You could proceed as follows (assuming $f\neq 0$): For every $n$, the sequence $x^n=(\text{sgn}(f(e_1)),\ldots,\text{sgn}(f(e_n)),0\ldots)$ is in $c_0$ and has norm $\leq 1$, so $\sum_{i=1}^n|f(e_i)|=|f(x^n)|\leq\Vert f\Vert$. This show that $(f(e_1),f(e_2),\ldots)\in \ell^1$.

Now, about you next question: First, verify that $c=c_0\oplus\mathbb{R}$. Then $c^*=c_0^*\oplus\mathbb{R}^*=\ell^1\oplus\mathbb{R}=\ell^1$, where the last isomorphism is given by $((x_1,x_2,x_3,\ldots,),\lambda)\mapsto(\lambda,x_1,x_2,x_3\ldots)$.

Luiz is right about your proof. To fix that, consider the following:

Let $f \in c_0^*$, then define $y = [y_1, y_2, ... ,y_n, ...]$ by $y_i = f(e_i), \forall i$. Observe that, if $f \in c_0^*$ then being a bounded linear functional we have that $\sup \{f(x) : x \in c_0^* \text{ and } \|x\| = 1 \} = M < \infty$. In particular, the limsup taken over the family of elements $\{x_n\}_{n=1}^{\infty}$, where $x_n = [\text{sgn}f(e_1) , \text{sgn}f(e_2), ... \text{sgn}f(e_n),\alpha, \alpha, ...]$ is finite. Thus $$\sum_{n=1}^{\infty} |y_i| = \limsup_{x_n} f(x_n) < \infty, $$ Consequently, $y \in \ell^{1}$ and we now have a method of defining a $y \in \ell^{1}$ from an $f \in c_{0}^*$. So that your mapping $\varphi : \ell^{1} \rightarrow c_{0}^*$ now has an inverse. Since you showed it preserves norms, and we now have it has an inverse it must be an isometry. Thus, you created an isometric embedding of $\ell^{1}$ onto $c_0^*$ in order to show the two vector spaces are "equal".

As for your later question, note that if you define $c_{\alpha} = \{ x \in \ell^{\infty} : x_n \rightarrow \alpha\}$ then this a vector space, if and only if $\alpha = 0$ for if $x,y \in c_{\alpha}$ then $x+y \in c_{2\alpha}$.