Let $\mathcal{C}_{[0,\infty)}$ denote all continuous functions: $[0, \infty) \to \mathbb{R}$. Let furthermore
$$ \mathcal{D}_1 = \bigcup_{t \in [0,1]} \{f \in \mathcal{C}_{[0, \infty)} \ | \ \text{$f$ is differentiable at $t$}\} $$
Show that $\mathcal{D}_1$ is a vector space.
It is quite clear that if $f \in \mathcal{D}_1$, then also $c f \in \mathcal{D}_1$ for any $c \in \mathbb{R}$ by the standard rules of diffentiability.
It is not clear to me that it is closed under addition since if we assume $f$ is differentiable in $t_1$ and $g$ is differentiable in $t_2$, how do we know that $f+g$ is differentiable anywhere?
The set $\mathcal{D}_1$ as defined in the question is not closed under addition.
Let $h$ a continuous but nowhere differentiable function, e.g. the Weierstraß function. Then for a fixed $a$ in the domain of $h$, the function $g \colon x \mapsto (x-a)\cdot h(x)$ is continuous, and differentiable at $a$, but nowhere else. By the continuity of $h$, we have
$$\lim_{x\to a} \frac{g(x) - g(a)}{x-a} = \lim_{x\to a} \frac{(x-a)h(x) - 0}{x-a} = \lim_{x\to a} h(x) = h(a)\,,$$
and if $g$ were differentiable at $y\neq a$, so would be $x\mapsto \frac{1}{x-a}\cdot g(x) = h(x)$, because $x \mapsto \frac{1}{x-a}$ is differentiable on $\mathbb{R}\setminus \{a\}$.
Thus, if $h$ is defined on $[0,\infty)$, the functions $f_1 \colon x \mapsto \bigl(x - \frac{1}{2}\bigr)h(x)$ and $f_2 \colon x\mapsto \bigl(\frac{2}{3} - x\bigr)h(x)$ both belong to $\mathcal{D}_1$, but $f_1 + f_2 \colon x \mapsto \frac{1}{6}h(x)$ doesn't.