Concluding $SPACE(n^2) \neq SPACE(n^7)$ from universal turing machine running time
I assume with ${\sf SPACE}$ you mean ${\sf DSpace}$. As usual for these types of hierarchy question you can use a diagonalization argument.

So build a machine $D$ that takes its input $w$ and then executes $M_u(\langle  w ,w \rangle)$. While executing $M_u$ it checks also how much space is used and how many steps the machine $M_u$ has done. Then the result of $D$ is the following:


if $D$ used more than $n^2$ space than reject,
if $D$ does more than $2^{cn^2}$ steps then accept (in this case the simulation detected a cycle if $\langle w \rangle$ would be a $n^2$ space bounded TM),
if none of the above occurred, then accept if $M_u$ rejected, and reject if $M_u$ accepted.


You can say that $D(w)$ anti-simulates the machine $M_u(\langle w ,w \rangle)$, if $\langle w \rangle$ is in ${\sf DSpace}(n^2)$. Clearly, $\langle D\rangle$ is in ${\sf DSpace}(n^7)$. However, it cannot be in ${\sf DSpace}(n^2)$. If it would be then you get a contradiction at $D(\langle D \rangle)$.