If $U$ and $W$ are subspaces of a finite dimensional inner product space $V$, show that
(a) If $U\subseteq W$, then $W^{\perp}\subseteq U^{\perp}$.
(b) $(U + W)^{\perp} = U^{\perp}\cap W^{\perp}$.
(c) $U^{\perp} + W^{\perp} \subset (U\cap W)^{\perp}$
MY ATTEMPT
(a) Let $\{u_{1},u_{2},\ldots,u_{m}\}$ be a basis for $U$. Since $U$ is a subspace of $W$, we can extend such basis to a basis of $W$ as $\{u_{1},\ldots,u_{m},u_{m+1},\ldots,u_{n}\}$. Consequently, if $\alpha\in W^{\perp}$, it is orthogonal to each vector $u_{k}$, where $1\leq k \leq n$. In particular, it is orthogonal to the set $\{u_{1},u_{2},\ldots,u_{m}\}$.
Thus $\alpha\in U^{\perp}$, that is to say, $W^{\perp}\subseteq U^{\perp}$.
(b) Let $\mathcal{B} = \{u_{1},u_{2},\ldots,u_{k}\}$ be a basis for $U\cap W$. Then we can extend it to a basis of $U$ and to a basis of $W$: $\mathcal{B}_{U} = \{u_{1},\ldots,u_{k},a_{k+1},\ldots,a_{m}\}$ and $\mathcal{B}_{W} = \{u_{1},\ldots,u_{k},b_{k+1},\ldots,b_{n}\}$. Hence a basis of $U+W$ is given by $\mathcal{B}_{U+W} = \mathcal{B}_{U}\cup\mathcal{B}_{W}$. Consequently, if $\beta$ is orthogonal to $U+W$, it is orthogonal to $\mathcal{B}_{U+W}$.
In particular, it is orthogonal to $\mathcal{B}_{U}$ and $\mathcal{B}_{W}$, from whence we conclude that $\beta\in U^{\perp}\cap W^{\perp}$.
Conversely, if $\beta$ is orthogonal to $U$ and orthogonal to $W$, it is orthogonal to $\mathcal{B}_{U}$ and $\mathcal{B}_{W}$, which means it is orthogonal to $\mathcal{B}_{U+W}$, and the claimed result holds.
(c) If $\alpha\in U^{\perp} + W^{\perp}$, then $\alpha = u + w$ where $u\in U^{\perp}$ and $w\in W^{\perp}$. Based on the notation in (b), we have that $u$ is orthogonal to $\mathcal{B}_{U}$ and $w$ is orthogonal to $\mathcal{B}_{W}$. Thus $u + w$ is orthogonal to $\mathcal{B}$, which is a basis for $U\cap W$. Hence $\alpha\in(U\cap W)^{\perp}$.
Could some one double-check my solution? Any other approach would be welcome.
EDIT
Based on the @user658409's comment, I have provided the following alternative solutions. If someone could double-check my solutions, I would appreciate.
(a) If $\alpha\in W^{\perp}$, then it is orthogonal to every element of $W$. Since $U\subseteq W$, $\alpha$ is orthogonal to $U$, from whence it results that $W^{\perp}\subseteq U^{\perp}$.
(b) Since $U\subseteq U + W$ and $W\subseteq U + W$, we have that $(U+W)^{\perp}\subseteq U^{\perp}$ and $(U+W)^{\perp}\subseteq W^{\perp}$. Thus $(U+W)^{\perp}\subseteq U^{\perp}\cap W^{\perp}$.
Conversely, if $\alpha\in U^{\perp}\cap W^{\perp}$, then $\alpha\in U^{\perp}$ and $\alpha\in W^{\perp}$. Thus $\alpha$ is orthogonal to any vector $u\in U$ and any vector $w\in W$. In particular, $\alpha$ is orthogonal to every $u + w\in U + W$, from whence the result holds.
(c) If $\alpha\in U^{\perp} + W^{\perp}$, then $\alpha = u + w$ where $u\in U^{\perp}$ and $w\in W^{\perp}$. Thus $u$ is orthogonal to every element of $U$ and $w$ is orthogonal to every element of $W$. In particular, $u$ is orthogonal to every element of $U\cap W$ and $w$ is orthogonal to every element of $U\cap W$. Thus $\alpha\in (U\cap W)^{\perp}$, as desired.
Best Answer
Your arguments are fine, but it could be simpler. It is probably that you used an alternative definition of the orthogonal complement, the traditional one is the following :
which in the case where $\textsf{V}$ is finite-dimensional, and $\beta$ is a basis for $\textsf{W}$, the latter reduces to $$\textsf{W}^\perp = \{ v \in \textsf{V} :\, \langle v,w \rangle = 0 \textrm{ for all } w \in \beta \}.$$ Notice that, in $(a)$, we don't need the finite dimensions: if $v \in W^\perp$, then $\langle v,w \rangle = 0$ for all $w\in W$, in particular for any vector of $U$. Hence $v\in U^\perp$.
Also, notice that, one inclusion of $(b)$ follows immediately of $(a)$: since $U \subset U + W$ and $W \subset U+W$, then $$(U + W)^\perp \subset U^\perp \quad \textrm{and} \quad (U + W)^\perp \subset W^\perp$$ and then $(U + W)^\perp \subset U^\perp \cap W^\perp$. The same for (c) since $$U \cap W \subset U \quad \textrm{and} \quad U \cap W \subset W$$ (you may have to use the fact that the sum of two subspaces is the smallest subspaces continaining both).