$1.$ $V=P_n(\mathbb{R}), $and $ W=\{p(x)\in P_n(\mathbb{R})\mid p(1)+p(2)+p(3)=0 \}$
$2.$ $V=M_{n\times n}(\mathbb{R}), $and $ W=\{A\in M_{n\times n}(\mathbb{R}) \mid A \text{ is not symmetric}\}$
$3.$ $V=M_{n\times n}(\mathbb{R}), $and $ W=\{A\in M_{n\times n}(\mathbb{R})\mid A \text{ is invertible}\}$
For the first one, I think let $p,q \in W$ and $c\in R$, then $(cp+q)(1)+(cp+q)(2)+(cp+q)(3)=c(p(1)+p(2)+p(3))+q(1)+q(2)+q(3)=0$. Therefore it is subspace of $V$.
For the second one, I think not symmetric can be represented as $a_{ij}\ne a_{ji} $ for some $j, i$. Then let $A, B\in W$ and $c\in R$, $ca_{ij}+b_{ij}-ca_{ji}+b_{ji}\ne 0$, $c(a_{ij}-a_{ji})+b_{ij}-b_{ji}\ne 0$. Here I saw a contradiction; what if $c$ is $0$?
For the last one, It can be written as $V=M_{n\times n}(\mathbb{R}), $and $ W=\{A\in M_{n\times n} (\mathbb{R}) \mid \det{A}\ne 0\}$. But how do you make it more explicit, as it's hard to prove this way?
Could anyone please correct my procedure?
Your proof for the first one is correct.
For the second one, look at $\left(\begin{matrix} 0 & 1\\ 0 & 0 \end{matrix}\right)+\left(\begin{matrix} 0 & 0\\ 1 & 0 \end{matrix}\right)$.
For the third, look at $\left(\begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right)+\left(\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix}\right)$.