$S$ and $T$ are subspaces of $\mathbb{R}^{n}$ and is defined as $S+T = \{v+w \mid v \in S \; and \; w \in T\}$. I need to show that $S+T$ is a subspace of $\mathbb{R}^{n}$.
Instinctively, $S+T$ is definitely inside $\mathbb{R}^{n}$ since $S \in \mathbb{R}^{n}$ and $T \in \mathbb{R}^{n}$. So the sum of any vectors in $S$ and $T$, although may not be a vector in both $S$ and $T$, ie: not inside $S \cap T$, it is still inside $\mathbb{R}^{n}$. But this is just my intuition and I want to prove it formally.
I thought I could use the subspace criteria $cv+dw$ to prove it.
Since $v \in S$ and $w \in T$, $cv \in S$ and $dw \in T$ where $c,d \in \mathbb{R}$.Then $cv+dw \in S+T \in \mathbb{R}^{n}$.
But somehow, I find what I've done isn't a very precise and convincing proof. How should I prove this more precisely?
Update of my attempt in the proof (Is this right?)
Let $\vec{v},\vec{w} \in S+T$, then $\vec{v}=\vec{s_1} + \vec{t_1}$ and $\vec{w}=\vec{s_2} + \vec{t_2}$ where $\vec{s_i} \in S, \; \vec{t_i} \in T$.
This implies that $\vec{v} = \vec{s_1}+\vec{t_1} \in S+T$. Let $c, d,r \in \mathbb{R}$, then $r\vec{v}=c\vec{s_1}+d\vec{t_1}$. Since $\vec{s_1}+\vec{t_1} \in S+T$, $c\vec{s_1}+d\vec{t_1} \in S+T \Rightarrow r\vec{v} \in S+T $.
Similarly, $\vec{w} = \vec{s_2}+\vec{t_2} \in S+T$. Let $j,k,s \in \mathbb{R}$, then $s\vec{w}=j\vec{s_2}+k\vec{t_2}$. Since $\vec{s_2}+\vec{t_2} \in S+T$, $j\vec{s_2}+k\vec{t_2} \in S+T \Rightarrow s\vec{w} \in S+T$.
Since $r\vec{v}, s\vec{w} \in S+T$, $r\vec{v}+s\vec{w} \in S+T$, hence $S+T$ is a subspace.