= Solution
Since $W$ is proper, $T$ is a nonzero <linear functional>. Choose $e\in X$ with $Te=1$, by rescaling any <vector> outside its <kernel>. For every $x\in X$,
$$
x=\underbrace{(x-(Tx)e)}_{\in W}+(Tx)e.
$$
If $ae\in W$, applying $T$ gives $a=0$. Consequently this is a <direct sum>, and the <quotient vector space> $X/W$ has <dimension> one. This computes the <codimension of a vector subspace>:
$$
\boxed{X=W\oplus\langle e\rangle,\qquad\operatorname{codim}_XW=1.}
$$
To prove boundedness without assuming <completeness>, use the <linear functional with closed kernel> criterion directly. Since $W$ is a <closed set> and $e\notin W$, the distance
$$
d=\inf_{w\in W}\|e-w\|_X
$$
is positive. If $Tx\ne0$, then $e-x/(Tx)\in W$, so
$$
d\leq\left\|\frac{x}{Tx}\right\|_X=\frac{\|x\|_X}{|Tx|}.
$$
For $Tx=0$ the resulting estimate is immediate. Therefore
$$
\boxed{|Tx|\leq d^{-1}\|x\|_X\quad(x\in X),\qquad T\in X^*.}
$$
Here $X^*$ denotes the <continuous dual space>.
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