Sums and Direct Sums of Subspaces My Question is twofold. 
First in Linear Algebra Done Right Axler states the following

The set $U_1+U_2+U_3+\dots +U_m$ is the smallest subspace which contains $U_1, U_2, U_3, \dots , U_m$ but he seems to state the proof in words only so i tried to proof it myself 

Is the following correct?
Proof. Given a vector space $V$ consider the following set.
$$A=\{S\mid\forall j\in\{1,2,3,\dots ,m\}(U_j\subseteq S)\}\tag{1}$$ where $S$ in $(1)$ is a subspace of $V$.
We seek to prove the following proposition $$\sum_{j=0}^m U_j\in A\ , \ \forall Y\in A \left(\sum_{j=0}^m U_j\subseteq Y\right) \tag{2}$$
It is fairly easy to see that $\sum_{j=0}^m U_j\in A$.
Let $Y$ be arbitrary and assume that $Y\in A$. Consider an arbitrary $x\in\sum_{j=0}^m U_j$ so that $x=\sum_{j=0}^m u_j$ where $$\forall j\in \{1,2,3,\dots , m\}(u_j\in U_j)\tag{3}$$
as a consequence of $(3)$ we see that $$\forall j\in \{1,2,3,\dots , m\}(u_j\in Y)\tag{4}$$
and since $Y$ is a subspace and therefore closed under addition it follows that $x\in Y$ since $x$ was arbitrary we may now conclude that $\sum_{j=0}^{m}U_j\subseteq Y$.
$\blacksquare$
Second Axler defines the notion of Direct Sum of subspaces as follows
The sum $U_1+U_2+\dots +U_m$ is the direct sum if each element of $U_1+U_2+\dots +U_m$ can be written in only one way as $u_1+u_2+\dots +u_m$ where each $u_j$ is in $U_j$ 
What does he mean by only one way?
 A: ‘Only one way’ means that if 
$$u_1+u_2+\dots+u_m=v_1+v_2+\dots+v_m,\quad\text{where}\; u_i, v_i\in U_i$$
then for each $i=1,\dots,m$, one has $u_i=v_i$.
Your proof other the first assertion is correct, but why do you think a proof  ‘only in words’ is not a real proof? I think that it's a better proof when possible, on the contrary.
A: I can't quite follow your argument for the first part. The idea for the proof is that the sum of $n$ subspaces $U_1+U_2+\cdots+U_n$ is the smallest subspace containing $U_1\cup U_2\cup\cdots\cup U_n$ in that it is created by taking the elements of $U_1\cup U_2\cup\cdots \cup U_n$ and adjoining all of their linear combinations.
As for your second question: A sum $U=U_1+U_2+\cdots+U_m$ is a direct sum if given any $u\in U$ written as $u=u_1+u_2+\cdots+u_n$ for $u_i\in U_i, 1\le i\le n,$ this representation is unique. I.e., if also
$ u=v_1+v_2+\cdots+v_n$ for $v_i\in U_i, 1\le i\le n$, we have $u_i=v_i.$
Alternatively, we can say that the map $\varphi:(U_1,\ldots, U_n)\to U_1+\cdots+U_n$ given by $\varphi(u_1,\ldots, u_n)=u_1+\cdots+u_n$ is injective. 
