[proofplan]
Since $Y$ is a proper closed subspace, we pick any $x \in X \setminus Y$ and note that $d := \operatorname{dist}(x, Y) > 0$ (closedness of $Y$ prevents $d = 0$). We then select $y_0 \in Y$ with $\|x - y_0\| < d/\theta$ and normalise the vector $x - y_0$ to obtain a unit vector whose distance to $Y$ is at least $\theta$.
[/proofplan]
[step:Choose $x \in X \setminus Y$ and show $\operatorname{dist}(x, Y) > 0$]
Since $Y \subsetneq X$, there exists $x \in X \setminus Y$. Define
\begin{align*}
d := \operatorname{dist}(x, Y) = \inf_{y \in Y} \|x - y\|.
\end{align*}
We claim $d > 0$. Suppose for contradiction that $d = 0$. Then there exists a sequence $(y_k)_{k \in \mathbb{N}}$ in $Y$ with $\|x - y_k\| \to 0$, so $y_k \to x$. Since $Y$ is closed, $x \in Y$, contradicting $x \in X \setminus Y$. Therefore $d > 0$.
[guided]
The entire construction hinges on the fact that a proper *closed* subspace stays a positive distance from any point outside it. This is precisely where the closedness hypothesis is consumed.
Since $Y \subsetneq X$, there exists $x \in X \setminus Y$. Define
\begin{align*}
d := \operatorname{dist}(x, Y) = \inf_{y \in Y} \|x - y\|.
\end{align*}
We claim $d > 0$. Suppose for contradiction that $d = 0$. Then for each $k \in \mathbb{N}$, we can find $y_k \in Y$ with $\|x - y_k\| < 1/k$, so $y_k \to x$ in the norm topology. Since $Y$ is closed, $Y$ contains all its limit points, hence $x \in Y$. This contradicts the choice $x \in X \setminus Y$. Therefore $d > 0$.
Why can we not take $\theta = 1$? Because the infimum $d = \inf_{y \in Y} \|x - y\|$ need not be attained — there may be no $y_0 \in Y$ achieving $\|x - y_0\| = d$. This is exactly the gap that forces $\theta < 1$: we can get within a factor of $\theta$ of the infimum, but not all the way. In finite-dimensional spaces, the infimum is always attained (closed bounded sets are compact), so one can take $\theta = 1$. In infinite-dimensional spaces, this generally fails.
[/guided]
[/step]
[step:Select an almost-closest point $y_0 \in Y$]
Since $\theta \in (0,1)$, we have $d/\theta > d$. By the definition of infimum, there exists $y_0 \in Y$ satisfying
\begin{align*}
d \le \|x - y_0\| < \frac{d}{\theta}.
\end{align*}
[guided]
We use the characterisation of the infimum: for any $\varepsilon > 0$, there exists $y \in Y$ with $\|x - y\| < d + \varepsilon$. We want $\|x - y_0\| < d/\theta$, so we need $d + \varepsilon \le d/\theta$, i.e., $\varepsilon \le d(1/\theta - 1)$. Since $\theta < 1$, we have $1/\theta - 1 > 0$, so such $\varepsilon$ exists. Choose any such $\varepsilon$ and take the corresponding $y_0 \in Y$. Then
\begin{align*}
d \le \|x - y_0\| < d + \varepsilon \le \frac{d}{\theta}.
\end{align*}
[/guided]
[/step]
[step:Normalise $x - y_0$ and verify the distance bound]
Define
\begin{align*}
x_\theta := \frac{x - y_0}{\|x - y_0\|}.
\end{align*}
Then $\|x_\theta\| = 1$ by construction. We verify $\operatorname{dist}(x_\theta, Y) \ge \theta$. For any $y \in Y$,
\begin{align*}
\|x_\theta - y\| = \left\|\frac{x - y_0}{\|x - y_0\|} - y\right\| = \frac{1}{\|x - y_0\|}\left\|x - y_0 - \|x - y_0\| \cdot y\right\| = \frac{1}{\|x - y_0\|}\left\|x - \underbrace{(y_0 + \|x - y_0\| \cdot y)}_{\in\, Y}\right\|.
\end{align*}
Since $y_0 \in Y$, $y \in Y$, and $Y$ is a subspace, $y_0 + \|x - y_0\| \cdot y \in Y$. Therefore
\begin{align*}
\|x_\theta - y\| \ge \frac{d}{\|x - y_0\|} > \frac{d}{d/\theta} = \theta.
\end{align*}
Taking the infimum over all $y \in Y$ gives $\operatorname{dist}(x_\theta, Y) \ge \theta$.
[guided]
Define
\begin{align*}
x_\theta := \frac{x - y_0}{\|x - y_0\|}.
\end{align*}
Since $x \notin Y$ and $y_0 \in Y$, we have $x - y_0 \neq 0$, so the normalisation is well-defined, and $\|x_\theta\| = 1$.
To show $\operatorname{dist}(x_\theta, Y) \ge \theta$, let $y \in Y$ be arbitrary. We compute
\begin{align*}
\|x_\theta - y\| &= \left\|\frac{x - y_0}{\|x - y_0\|} - y\right\| = \frac{1}{\|x - y_0\|} \left\|x - y_0 - \|x - y_0\| \cdot y\right\|.
\end{align*}
Write $\tilde{y} := y_0 + \|x - y_0\| \cdot y$. Since $Y$ is a linear subspace containing both $y_0$ and $y$, and since $\|x - y_0\|$ is a scalar, $\tilde{y} \in Y$. Therefore
\begin{align*}
\|x_\theta - y\| = \frac{\|x - \tilde{y}\|}{\|x - y_0\|} \ge \frac{\operatorname{dist}(x, Y)}{\|x - y_0\|} = \frac{d}{\|x - y_0\|}.
\end{align*}
Now we use the bound $\|x - y_0\| < d/\theta$ from the previous step:
\begin{align*}
\|x_\theta - y\| > \frac{d}{d/\theta} = \theta.
\end{align*}
Since this holds for every $y \in Y$, we conclude $\operatorname{dist}(x_\theta, Y) = \inf_{y \in Y} \|x_\theta - y\| \ge \theta$.
The key mechanism: every element of $Y$ that might be close to $x_\theta$ corresponds, after undoing the normalisation, to an element of $Y$ close to $x$. But $x$ is at distance $d > 0$ from $Y$, and we chose $y_0$ so that $\|x - y_0\|$ exceeds $d$ by at most a factor of $1/\theta$. The ratio $d / \|x - y_0\|$ therefore remains at least $\theta$.
[/guided]
[/step]
