Given a linear operator $T$ acting on a vector space $V$ an invariant subspace of $T$ is a subspace $W$ of $V$ such that $T(W)\subseteq W$.
So, We always have two invariant subspaces: $Ker (T)$ and $Range(T)$.
In this case we can easily see that $Ker (T)=\{0\}$ and $Range(T)=\mathbb{R}^2$ that are two trivial invariant subspaces of dimension $0$ and $2$.
Invariant subspaces of dimension $1$ are spanned by a vector $x$ such that $T(x)=\lambda x$, i.e. eigenvectors of the matrix that represents $T$:
$$
\begin{bmatrix}
1&1\\
0&1
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
= \lambda
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
$$
The matrix is exactly in Jordan canonical form, so we see immediately that it has an eigenvalue $\lambda=1$ with algebraic multiplicity $2$ and only one proper eigenvector $[1,0]^T$ that span the eigenspace $W=\{[k,0]^T\}$, so this is an invariant subspace of dimension $1$. And there are no other invariant subspaces.
About the last question: edit this answer and look how I've write the matrices.
Without using Jordan form.
An eigenvector of the eigenvalue $\lambda=1$ satisfies the equation:
$$
\begin{bmatrix}
1&1\\
0&1
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
= 1
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
$$
that is :
$$
\begin{cases}
x_1+x_2=x_1\\
x_2=x_2
\end{cases}
$$
that has solution $x_2=0 \;\land\;x_1=k \, \forall k \in \mathbb{R}$. This means that all vectors of the form $[k,0]^T$ are solutions, i.e. are elements of the eigenspace, and as an eigenvector we can chose $[1,0]^T$.
