You can prove it without heavy machinery, but a bit of computation, using distance-minimising sequences.
Suppose $V \le H$ is closed and proper. Choose any point $x_0 \in H \setminus V$. Let
$$r = \inf_{x \in V} \|x - x_0\| > 0,$$
since $V$ is closed. Suppose $(x_n)_{n \ge 1} \in V$ such that $\lim_{n \to \infty} \|x_n - x_0\| = r$. I claim that $(x_n)$ is Cauchy.
Fix $\varepsilon > 0$. Let
$$\delta = \sqrt{r^2 + \frac{\varepsilon}{4}} - r > 0.$$
Since $\|x_n - x_0\| \to r$, there exists some $N$ such that
$$n \ge N \implies r \le \|x_n - x_0\| < r + \delta = \sqrt{r^2 + \frac{\varepsilon}{4}}.$$
Suppose $m, n \ge N$. Parallelogram law states that:
$$\|(x_m - x_0) - (x_n - x_0)\|^2 + \|(x_m - x_0) + (x_n - x_0)\|^2 = 2\|x_m - x_0\|^2 + 2\|x_n - x_0\|^2,$$
or equivalently,
$$\|x_m - x_n\|^2 = 2\|x_m - x_0\|^2 + 2\|x_n - x_0\|^2 - 4\left\|\frac{x_m + x_n}{2} - x_0\right\|^2.$$
Note that $\frac{x_m + x_n}{2} \in V$, so
$$\left\|\frac{x_m + x_n}{2} - x_0\right\|^2 \ge r^2,$$
by definition of $r$. Therefore,
$$\|x_m - x_n\|^2 < 2\left(r^2 + \frac{\varepsilon}{4}\right) + 2\left(r^2 + \frac{\varepsilon}{4}\right) - 4r^2 = \varepsilon.$$
Thus, $(x_n)$ is Cauchy. It has a limit $x_\infty$, by completeness. I claim that $x_\infty - x_0 \in V^\perp \setminus \{0\}$.
First,
\begin{align*}
\|x_\infty - x_0\| &= \left\|\lim_{n \to \infty} x_n - x_0\right\| \\
&= \lim_{n \to \infty} \|x_n - x_0\| \\
&= r > 0,
\end{align*}
so $x_\infty - x_0 \neq 0$.
To see that $x_\infty - x_0 \in V^\perp$, fix $v \in V \setminus \{0\}$, and suppose $t \in \Bbb{C}$ (or just $\Bbb{R}$, if we're in the reals) to be decided shortly. Then $x_\infty + tv \in V$, so
$$\|x_\infty + tv - x_0\| \ge r = \|x_\infty - x_0\|.$$
Then
\begin{align*}
0 &\le \|x_\infty - x_0 + tv\|^2 - \|x_\infty - x_0\|^2 \\
&= |t|^2\|v\|^2 - 2\operatorname{Re}(t \langle x_\infty - x_0, v \rangle).
\end{align*}
In particular, let
$$t = \frac{\langle v, x_\infty - x_0\rangle}{\|v\|^2},$$
so that the above becomes:
$$0 \le -\frac{|\langle x_\infty - x_0, v\rangle|^2}{\|v\|^2},$$
which implies $\langle v, x_\infty - x_0\rangle = 0$ for all $v \in V \setminus \{0\}$ (and hence all $v \in V$). Thus $x_\infty - x_0 \in V^\perp$.
Best Answer
A closed subspace of Banach space is always a Banach space and the orthogonal complement of any set is always closed. So the only thing you have to check is that $T'$ is a bijection.
If $T'x=0$ for some $x \in \mathcal N(T)^{\perp}$ then $x \in \mathcal N(T) \cap \mathcal N(T)^{\perp} =\{0\}$ so $x=0$. So $T'$ is injective.
Let $y \in \mathcal R{(T)}$. Then $y=Tx$ for some $x$ and we can write $x=x_1+x_2$ where $x_1 \in \mathcal N(T)$ and $x_2 \in \mathcal N(T)^{\perp}$. This gives $y=T'x_2$ so $T'$ is surjective.