The sum of subspaces is the smallest subspace containing all the summands In Axler's Linear Algebra Done Right the theorem given is 

Suppose $U_1,\ldots, U_m$ are subspaces of $V$. Then $U_1+\cdots+ U_m$ is the smallest subspace of $V$ containing $U_1,\ldots, U_m.$

I can see that the sum will be a subspace of $V$. What I don't understand is the following paragraph:
"Clearly $U_1,\ldots, U_m$ are all contained in $U_1+\cdots+ U_m$ (to see this consider sums $u_1 + \cdots+u_m$ where all except one of the $u$s is $0$). Conversely every subspace of V containing $U_1,\ldots, U_m$ must contain $U_1+\cdots+ U_m$ (because subspaces must contain all finite sums of their elements). Thus, $U_1+\cdots+ U_m$ is the smallest subspace of $V$ containing $U_1,\cdots, U_m."$
This question was addressed here: Misunderstanding in the proof that the sum of subspaces is the smallest containing subspace. but the answer wasn't clear enough for me.
I can see how the first part is true, that  $U_1,\ldots, U_m$ are all contained in $U_1+\cdots+ U_m$ and the second part, $U_1,\ldots, U_m$ must contain $U_1+\cdots+ U_m$ but I don't understand how together they prove the theorem.
Can someone please take the trouble of giving an intuitive explanation, or a geometric one if it is possible? Thanks a lot. 
 A: 
[…] and second part, $U_1, \dotsc, U_m$ must contain $U_1 + \dotsb + U_m$.

This is wrong. Consider for example $V = \mathbb{R}^2$ and the subspaces
$$
 U_1 = \{(x,0) \mid x \in \mathbb{R}\}
 \quad\text{and}\quad
 U_2 = \{(0,y) \mid y \in \mathbb{R}\}.
$$
Then $U_1 + U_2 = \mathbb{R}^2$, but neither $U_1$ nor $U_2$ contain $\mathbb{R}^2$.
The statement that we really want is the following:

$U_1 + \dotsb + U_n$ is a subspace containing $U_1, \dotsc, U_n$, and if $W \subseteq V$ is any subspace with $U_1, \dotsc, U_n \subseteq W$, then we already have $U_1 + \dotsb + U_n \subseteq W$.

You already understand the first part of this statement, that $U_1 + \dotsb + U_n$ is a subspace containing $U_1, \dotsc, U_n$.
For the other part suppose $W \subseteq V$ is some subspace containing $U_1, \dotsc, U_n$. For all $u_1 \in U_1, \dotsc, u_n \in U_n$, we then have $u_1, \dotsc, u_n \in W$. Because $W$ is a subspace it follows that also $u_1 + \dotsb + u_n \in W$ (because $W$ is closed under finite sums). Therefore 
$$
 U_1 + \dotsb + U_n
 = \{u_1 + \dotsb + u_n \mid u_1 \in U_1, \dotsc, u_n \in U_n\}
 \subseteq W.
$$
A: I'm not sure whether it is clear to me what you are asking, but this is the classical case where you show that two sets $X,Y$ are equal by showing that both $X\subset Y $ and $Y\subset X$ hold true.
A: Proof by contradiction:
Let S be a smaller subspace than U1+U2+...+Um that contains U1,U2,...,Um.
S contains U1,U2,...,Um and because it is a subspace it must be closed under addition therefore U1+U2+...+Um has to be contained within S. In contradiction to our assumption that it is smaller than U1+U2...+Um
Credit's due to my sister Hadar.
A: Apologies for adding another answer years later, but I just went through the same dilemma while going through this book and think this will help support Jendrik's answer. I'll focus on the second part of the proof as that's where most of the confusion lies.
What we're trying to prove:

Suppose $U_1,…,U_m$ are subspaces of $V$. Then $U_1+⋯+U_m$ is the smallest subspace of $V$ containing $U_1,…,U_m$.

We will show that each subspace $U_1,…,U_m \subseteq U_1+⋯+U_m$. From the book:

Clearly $U_1,…,U_m$ are all contained in $U_1+⋯+U_m$ (to see this consider sums $U_1+⋯+U_m$ where all except one of the us is 0).

Now, we will prove that every subspace in $V$ that contains $U_1,…,U_m$ will also contain $U_1+⋯+U_m$.
Again, from the book:

Conversely every subspace of $V$ containing $U_1,…,U_m$ must contain $U_1+⋯+U_m$ (because subspaces must contain all finite sums of their elements).

Vectors spaces are closed under addition, meaning the sum of two elements from the vector space must also be in the vector space. So, we can take the sum of elements $u_1 + u_2 + ... + u_m \text{ where } u_i \in U_i$ and get any element from $U_1+⋯+U_m$. So, any subspace containing $U_1,…,U_m$ also contains $U_1+⋯+U_m$.
Why does this mean $U_1+⋯+U_m$ is smallest possible subspace containing $U_1,…,U_m$ ? Let $T$ be a subspace that contains $U_1,…,U_m$. Since $T$ contains $U_1,…,U_m$, we know that it must also contain $U_1+⋯+U_m$. Therefore, $U_1+⋯+U_m \subseteq T$ and $U_1,…,U_m \subseteq U_1+⋯+U_m \subseteq T$.
