Let $(E, |\cdot|)$ be a Banach space. Then we have
Riesz's lemma Let $L$ be closed proper subspace of $E$. Then for each $\varepsilon \in (0, 1)$, there is $x \in E$ such that $|x|=1$ and $d(x, L) := \inf_{y\in L} |x-y| \ge \varepsilon$.
I would like to ask if the following statement holds, i.e.,
Let $G,L$ be two non-trivial subspaces of $E$ such that $L$ is closed and $G \cap L = \{0\}$. Then for each $\varepsilon \in (0, 1)$, there is $x \in G$ such that $|x|=1$ and $d(x, L) := \inf_{y\in L} |x-y| \ge \varepsilon$.
Thank you so much for your elaboration!
As Anne and Ryszard mentioned in comments, a counter-example would be two lines passing through the origin in $E:=\mathbb R^2$. In particular, Ryszard's counter-example is as follows. Let $G :=\{(x, x) : x \in \mathbb R\}$ and $L:=\{(x, 0) : x \in \mathbb R\}$. If $u \in G$ and $|u|=1$, then $u= \pm \frac{1}{\sqrt 2} (1, 1)$. Then $$ d(u, L) = \inf_{x\in \mathbb R} \sqrt{ \left | \frac{1}{\sqrt 2}-x \right |^2 + \left | \frac{1}{\sqrt 2}-0 \right |^2} \ge \frac{1}{\sqrt 2}. $$
If we take $\varepsilon \in (0, \frac{1}{\sqrt 2})$, then the statement fails.