Apply Theorem TSS.
First, the zero vector of $M_{nn}$ is the zero matrix, $\zeromatrix$, whose entries are all zero (Definition ZM). This matrix then meets the condition that $\matrixentry{\zeromatrix}{ij}=0$ for $i>j$ and so is an element of $UT_n$.
Suppose $A,B\in UT_n$. Is $A+B\in UT_n$? We examine the entries of $A+B$ “below” the diagonal. That is, in the following, assume that $i>j$.
\begin{align*}
\matrixentry{A+B}{ij}
&=\matrixentry{A}{ij}+\matrixentry{B}{ij}&&\knowl{./knowls/definition.MA.knowl}{\text{Definition MA}}\\
&=0 + 0&&\text{$A,B\in UT_n$}\\
&=0
\end{align*}
which qualifies $A+B$ for membership in $UT_n$.
Suppose $\alpha\in\complex{}$ and $A\in UT_n$. Is $\alpha A\in UT_n$? We examine the entries of $\alpha A$ “below” the diagonal. That is, in the following, assume that $i>j$.
\begin{align*}
\matrixentry{\alpha A}{ij}
&=\alpha\matrixentry{A}{ij}&&\knowl{./knowls/definition.MSM.knowl}{\text{Definition MSM}}\\
&=\alpha 0&&\text{$A\in UT_n$}\\
&=0
\end{align*}
which qualifies $\alpha A$ for membership in $UT_n$.
Having fulfilled the three conditions of Theorem TSS we see that $UT_n$ is a subspace of $M_{nn}$.