Show that there is no non-trivial reducing subspace of the unilateral shift $T:l^2\to l^2$ given by $T((x_1,x_2,...))=(0,x_1,x_2,...)$.
So suppose there is a non-trivial reducing subspace $M$ then we know by definition $T(M)\subset M$ and $T(M^{\perp})\subset M^{\perp}$. Then there exist non-zero $x=(x_1,x_2,...)\in M$ and non-zero $y=(y_1,y_2,...)\in M^{\perp}$.
Using $\langle x,y\rangle=0,\langle Tx,y\rangle = 0, \langle x,Ty\rangle =0$ we get the following equations:
$$\sum_n x_ny_n=0,\sum_nx_ny_{n+1}=0,\sum_nx_{n+1}y_n=0$$
How to show from this that such non-zero $x,y$ cannot exist?
Let $M$ be such a non-trivial reducing subspace.
We can assume $M$ to be a closed linear subspace: Let $x\in \bar M$. Then there is a sequence $(x_k)$ with $x_k\to x$ and $x_k\in M$, then $Tx_k\in M$, $Tx_k\to x$, which implies $Tx\in \bar M$. In addition, $(\bar M)^\perp = M^\perp$.
Then $l^2$ is a direct (orthogonal) sum of $M$ and $M^\perp$, we have for the first unit vector $$ e_1 = x + y $$ with $x\in M$, $y\in M^\perp$. Taking the inner product with $x$ and $y$ yields $\langle e_1,x\rangle=\|x\|^2$ and $\langle e_1,y\rangle=\|y\|^2$. Since not both of $x$ and $y$ are zero, this implies $e_1=x$ or $e_1=y$.
This implies $e_1\in M$ or $e_1\in M^\perp$, or in other words, $e_1\in V$, where $V$ is an invariant subspace of $T$, i.e., $TV\subset V$. This implies that all unit vectors $e_k$ are in $V$. The (closed) subspace $V$ thus contains the closed linear hull of these $(e_k)$, hence $V=l^2$.
This proves that $M=l^2$ or $M^\perp=l^2$, which is a contradiction to non-triviality.