Let $C \subseteq \mathbb R^n$, and $C \subseteq D$ where $D$ is affine.
Show that $\text{aff }C \subseteq D$
What I tried:
Take $y \in \text{aff }C$. Then $y = \sum_{i=1}^{k}t_ix_i$ where $x_i \in C$ and $\sum_{i=1}^{k}t_i = 1$.
We wish to show that $y \in D$. Once we show that, we're done.
Now, since $C \subseteq D$, then $x_i \in D$ as well. So $y = \sum_{i=1}^{k}t_ix_i$ is an affine combination of elements in $D$ too.
That's all well and good, but I want to reach the form $y = \lambda w_1 + (1-\lambda) w_2$ where $w_1, w_2 \in D$. Then I would know that $y \in D$, because $D$ is affine.
Write $t_2 = 1-t_1 - t_3 -\dots -t_k$.
Then $y = t_1x_1 + (1-t_1 - t_3 - \dots -t_k)x_2 + t_3x_3+\dots t_kx_k = \\t_1x_1+(1-t_1)x_2+(-t_3-t_4-\dots -t_k)x_2+t_3x_3+\dots t_kx_k$
Since $D$ is affine, $t_1x_1 + (1-t_1)x_2 \in D$
I'm unsure how to continue now or if this is even the right direction
I realize this has been asked before, but the answer there only works if $C$ is known to be a linear subspace. We don't know that here.
Note that if we choose an element $x_0 \in C$ and define $$ \hat C = \{x - x_0 : x \in C\}; \quad \hat D = \{x - x_0 : x \in D\}, $$ then it suffices to apply the result on linear subspaces to $\hat C \subseteq \hat D$.
That being said, to complete your proof it suffices to apply the following:
Proof: Proceed inductively. Suppose that we know that any affine combination of $n \geq 2$ elements is necessarily an element of $D$. Now, consider $\sum_{i=1}^{n+1} t_i x_i$. It cannot be the case that $t_i = 1$ for all $i$, so without loss of generality suppose that $t_{n+1} \neq 1$. We have $$ \sum_{i=1}^{n+1} t_i x_i = \sum_{i=1}^{n} t_i x_i + t_{n+1}x_{n+1} = (1-t_{n+1})\left(\sum_{i=1}^{n} \frac{t_i}{1 - t_{n+1}} x_i\right) + t_{n+1}x_{n+1} $$ So that $\sum_{i=1}^{n+1} t_i x_i$ is an affine combination of two elements of $D$ and is therefore in $D$.