@Vini provides a good solution. Another presentation1 follows.
Definition of the singular value decomposition:
$$
\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
$$
$$
%
\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}, \quad
%
\mathbf{U} \in \mathbb{C}^{m\times m}, \quad
%
\mathbf{V} \in \mathbb{C}^{n\times n}, \quad
%
\Sigma \in \mathbb{R}^{m\times n}, \quad
%
\mathbf{S} \in \mathbb{R}^{\rho\times\rho}
%
$$
The domain matrices are unitary:
$$
\begin{align}
\mathbf{U}^{*} \mathbf{U} &= \mathbf{U} \, \mathbf{U}^{*} = \mathbf{I}_{m} \\
\mathbf{V}^{*} \mathbf{V} &= \mathbf{V} \, \mathbf{V}^{*} = \mathbf{I}_{n}
\end{align}
$$
For this problem $m=n=\rho=2$ and both nullspaces are trivial.
1 Compute product matrix
$$
\mathbf{A}^{*} \mathbf{A} =
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 0 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{cc}
2 & 1 \\
1 & 1 \\
\end{array}
\right]
$$
2 Resolve eigensystem
$$
\lambda \left( \mathbf{A}^{*} \mathbf{A} \right) =
\frac{1}{2} \left( 1 \pm \sqrt{5} \right)
$$
The choice of eigenvectors involves a choice of sign. This is the origin of the sign ambiguities of the SVD. The signs are chosen here with the domain matrix $\mathbf{V}$ and they will imprint in the codomain matrix $\mathbf{U}$.
$$
v_{1} =
\left[
\begin{array}{c}
\frac{1}{2} \left(1+\sqrt{5}\right) \\
1
\end{array}
\right], \quad
%
v_{2} =
\left[
\begin{array}{c}
\frac{1}{2} \left(1 - \sqrt{5}\right) \\
1
\end{array}
\right]
$$
3 Assemble $\Sigma$
The singular values of $\mathbf{A}$ are the square roots of the nonzero eigenvalues of the product matrix:
$$
\sigma
= \sqrt{\lambda \left( \mathbf{A}^{*} \mathbf{A} \right)}
= \sqrt{\frac{1}{2} \left( 1 \pm \sqrt{5} \right)}.
$$
The eigenvalues are ordered:
$$
\sigma_{1} \ge \sigma_{2} \gt 0
$$
The diagonal matrix of singular values is
$$
\mathbf{S} =
\left[
\begin{array}{cc}
\sigma_{1} & 0 \\
0 & \sigma_{2} \\
\end{array}
\right]
%
=
%
\frac{1}{2}
\left[
\begin{array}{cc}
1 + \sqrt{5} & 0 \\
0 & 1 - \sqrt{5} \\
\end{array}
\right].
$$
in the general case this matrix is embedded in a sabot matrix which guarantees conformability:
$$
\Sigma =
\left[
\begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} \\
\end{array}
\right]
\in
\mathbb{R}^{m\times n}.
$$
Because both nullspace are trivial in this example, $\Sigma = \mathbf{S}.$
4 Assemble domain matrix $\mathbf{V}$
$$
\begin{align}
\mathbf{V} =
%
\left[
\begin{array}{cc}
\frac{v_{1}} {\lVert v_{1} \rVert} & \frac{v_{2}} {\lVert v_{2} \rVert}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{cc}
\frac{1}{\sqrt{ \frac{1}{2} \left(5+\sqrt{5}\right)}}
\left[
\begin{array}{c}
\frac{1}{2} \left(1+\sqrt{5}\right) \\
1
\end{array}
\right] &
\frac{1}{\sqrt{ \frac{1}{2} \left(5-\sqrt{5}\right)}}
\left[
\begin{array}{c}
\frac{1}{2} \left(1-\sqrt{5}\right) \\
1
\end{array}
\right]
\end{array}
\right] \\[5pt]
%
&=
\left[
\begin{array}{cc}
\sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} & -\sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} \\
\sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} & \sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} \\
\end{array}
\right]
%
\end{align}
%
$$
5 Construct codomain matrix $\mathbf{U}$
From the definition
$$
\mathbf{A} = \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*}
\qquad \Rightarrow \qquad
\mathbf{A} \, \mathbf{V} = \mathbf{S} \,\mathbf{U}
$$
Therefore
$$
\mathbf{U}_{k} = \sigma_{k}^{-1} \left[ \mathbf{A} \, \mathbf{V} \right]_{k}
$$
After some algebra,
$$
\mathbf{U} =
\left[
\begin{array}{rr}
\sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} & \sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} \\
\sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} & -\sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} \\
\end{array}
\right]
$$
Verify decomposition
$$
\begin{align}
%
\mathbf{A} &= \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*} \\
% A
\left[
\begin{array}{cc}
1 & 1 \\
1 & 0 \\
\end{array}
\right]
%
&=
% U
\left[
\begin{array}{cc}
\sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} & \sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} \\
\sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} & -\sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} \\
\end{array}
\right]
% S
\frac{1}{2}
\left[
\begin{array}{cc}
1 + \sqrt{5} & 0 \\
0 & 1 - \sqrt{5} \\
\end{array}
\right]
% V*
\left[
\begin{array}{rr}
\sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)} & \sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} \\
-\sqrt{\frac{1}{10} \left(5-\sqrt{5}\right)} & \sqrt{\frac{1}{10} \left(5+\sqrt{5}\right)}
\end{array}
\right]
%
\end{align}
$$
Compare to diagonalization
$$
\mathbf{A} = \mathbf{P} \, \mathbf{D} \, \mathbf{P}^{-1}
$$
The diagonal matrix $\mathbf{D}$ uses the eigenvalues of $\mathbf{A}$, diagonalizer $\mathbf{P}$ is constructed from eigenvectors of $\mathbf{A}$.
$$
%
\mathbf{P} =
\frac{1}{2}
\left[
\begin{array}{cc}
\left(1+\sqrt{5}\right) & 2 \\
\left(1-\sqrt{5}\right) & 2 \\
\end{array}
\right], \quad
%
\mathbf{D} =
\frac{1}{2}
\left[
\begin{array}{cc}
1+\sqrt{5} & 0 \\
0 & 1-\sqrt{5} \\
\end{array}
\right], \quad
%
\mathbf{P}^{-1} =
\frac{1}{2\sqrt{5}}
\left[
\begin{array}{cc}
2 & -2 \\
-1+\sqrt{5} & 1+\sqrt{5} \\
\end{array}
\right]
%
$$
