Prove the supremum of the set of affine functions is convex

Let $(g_i)_{i\in I}$ be a family of convex functions on a convex compact set $\Omega\subseteq \mathbb{R}^d$. We will show that the sup of this family is convex. We will use the standard definition of convexity.

Let $g:=\sup_{i\in I} g_i$.

Take $x,y\in\Omega$ and $t\in[0,1]$.

Fix $i\in I$. Since $g_i$ is convex and bounded above by $g$, we have $$ g_i(tx+(1-t)y)\leq tg_i(x)+(1-t)g_i(y)\leq tg(x)+(1-t)g(y). $$ Since the latter holds for every $i\in I$, we can take the sup and find $$ g(tx+(1-t)y)\leq tg(x)+(1-t)g(y). $$

This holds for every $x,y\in \Omega$ and every $t\in[0,1]$. So $g$ is convex.

Now every affine function $f_i$ is convex, so the result follows from the general case above.

Geometrically? A function is convex iff its epigraph is convex. See here for a definition of the epigraph. It is clear that the epigraph of $\sup g_i$ is the intersection of the epigraphs of all the $g_i$. Now the intersection of convex sets is convex, which yields a more geometric proof of the statement above.

I can't comment because I don't have enough reputation, hence posting as an answer. I don't think $f \in C^1(\Omega)$. Easiest example is take $\Omega = [0,1]$, $I = \{1,2\}$. $f_1(x) = \frac12$ and $f_2(x) = x$. Then, $f(x) = \frac12$ if $x \le \frac12$ and $x$ if $x > \frac12$. In particular, at $\frac12$ there is a kink.