Show that $U_1 + U_2 + ... + U_k = \operatorname{span}(U_1 \cup U_2 \cup ... \cup U_k)$ Let us use induction.
1) Base case $k=2$
$U_1 + U_2 = span (U_1 \cup U_2)$, which I understand how to prove is OK (Sum of two subspaces is equal to the span of their union)
2) Induction hypothesis
We assume that the following statement holds
$$U_1 + U_2 + ... + U_{k-1} = \operatorname{span} (U_1 \cup U_2 \cup \ ... \ \cup U_{k-1})$$
3) Induction step
$$U_1 + U_2 + ... + U_k = \underbrace{U_1 + U_2 + ... + U_{k-1}}_{=\operatorname{span} (U_1 \cup U_2 \cup \ ... \ \cup U_{k-1})} + U_k = \ ? $$
But my issue is that I do not see how to move from $\operatorname{span} (U_1 \cup U_2 \cup \ ... \ \cup U_{k-1})+U_k$
 A: You don't really need induction.
If $x = u_1 + \ldots + u_n$ is a representation of a vector in $\mathrm{U}_1 + \ldots + \mathrm{U}_n$ then $x \in \mathrm{span}(\mathrm{U}_1 \cup \ldots \cup \mathrm{U}_n)$ (since the span is closed under linear combinations by definition). Reciprocally, if $x$ belongs to the span, that means that $x$ is a linear combination of $u_1, \ldots, u_n \in \mathrm{U}_1 \cup \ldots \cup \mathrm{U}_n,$ and since each $\mathrm{U}_i$ is closed under linear operations (viz. addition and multiplication by scalar), we may assume each $u_i$ belongs to $\mathrm{U}_i$ and we are done (by definition of a sum of vector spaces).
Assuming you are very new, let me expand on a point made above. I am going to carefully prove that if $x \in \mathrm{span}(\mathrm{U}_1 \cup \ldots \cup \mathrm{U}_n)$ then $x \in \mathrm{U}_1 + \ldots + \mathrm{U}_n.$ By definition, $x \in \mathrm{span}(\mathrm{U}_1 \cup \ldots \cup \mathrm{U}_n)$ signifies that there are vectors $u_1, \ldots, u_r \in \mathrm{U}_1 \cup \ldots \cup \mathrm{U}_n$ such that $x = \sum\limits_{j = 1}^r \beta_j u_j.$ Construct the following sets:

*

*$\mathrm{K}_1$ the set of indices $j$ such that $u_j \in \mathrm{U}_1,$

*$\mathrm{K}_2$ is the set of indices $j$ not in $\mathrm{K}_1$ such that $u_j \in \mathrm{U}_2,$

*and so on (I guess this uses induction, oops!).

Then, the $\mathrm{K}_i$ are disjoint and $x = \sum\limits_{i = 1}^n \sum\limits_{j \in \mathrm{K}_j} \beta_j u_j$ and by the construction of the $\mathrm{K}_i,$ $\sum\limits_{j \in \mathrm{K}_i} \beta_j u_j \in \mathrm{U}_i$ and we are done.
A: Note that $span(S)$ is the unique subspace satisfying the following property: for all subspaces $Q$, $S \subseteq Q$ iff $span(S) \subseteq(Q)$.
The base case is actually $k = 0$, not $k = 2$. For the base case, we see that the empty sum, which is the zero space, is equal to the span of the empty union, that is, the span of the empty set. This is because it is always true that $\emptyset \leq Q$, and whenever $Q$ is a subspace, it is always true that $0 \subseteq Q$. So $0 = span(\emptyset)$.
Now for the inductive step. We see that $(x_1 + ... + x_n) + x_{n + 1} = span(x_1 \cup ... \cup x_n) + x_{n + 1}$. You have shown that for all subspaces $a$ and $b$, $a + b = span(a \cup b)$. So we see that $span(x_1 \cup ... \cup x_n) + x_{n + 1} = span(span(x_1 \cup ... \cup x_n) \cup x_{n + 1})$.
Finally, we see that for any subspace $Q$, we have $span(span(x_1 \cup ... \cup x_n) \cup x_{n + 1}) \subseteq Q$ iff $span(x_1 \cup ... \cup x_n) \cup x_{n + 1} \subseteq Q$ iff ($span(x_1 \cup ... \cup x_n) \subseteq Q$ and $x_{n + 1} \subseteq Q$) iff ($x_1 \cup ... \cup x_n \subseteq Q$ and $x_{n + 1} \subseteq Q$) iff $x_1 \cup ... \cup x_n \cup x_{n + 1} \subseteq Q$ iff $span(x_1 \cup ... \cup x_n \cup x_{n + 1}) \subseteq Q$.
Therefore, $span(span(x_1 \cup ... \cup x_n) \cup x_{n + 1}) = span(x_1 \cup ... \cup x_n \cup x_{n + 1})$.
In fact, this is a special case of left adjoints preserving colimits. We can show that $+$ is the join operator for subspaces. Since $span$ is a left adjoint, it preserves joins. So this means that in general, $span(S_1) + ... + span(S_n) = span(S_1 \cup ... \cup S_n)$. In the case where $U_i$ is a subspace, we see that $U_i = span(U_i)$ and thus $U_1 + ... + U_n = span(U_1) + ... + span(U_n) = span(U_1 \cup ... \cup U_n)$. There is thus little need for induction here.
Elaborating on why the last bit is true without category theory: we see that for all subspaces $x_1, ..., x_n$ and $Q$, we have $x_1 + ... + x_n \subseteq Q$ iff for all $i$, $x_i \subseteq Q$. So therefore, we have $span(s_1) \cup ... \cup span(s_n) \subseteq Q$ iff (for all $i$, $span(s_i) \subseteq Q$) iff (for all $i$, $s_i \subseteq Q$) iff ($s_1 \cup ... \cup s_n \subseteq Q$) iff ($span(s_1 \cup ... \cup s_n) \subseteq Q$).
