Let $\lbrace e_1,e_2,e_3\rbrace$ be the standard basis on $\mathbb{R^3}$.
Let $I_1$ be the line segment between $e_1$ and $e_3$, $I_2$ the line segment between $0$ and $e_3$, $I_3$ the line segment between $e_2$ and $e_3$ and $T$ the triangle with vertices $0, e_1, e_2$.
Define $E:=I_1 \cup I_2 \cup I_3 \cup T$ with $E \subset \mathbb{R^3}$
How to compute the Hausdorff measure $\mathcal{H^2}(E)$ and conclude that dim$_\mathcal{H}(E)=2$?
I tried to use that there is a map $S: \mathbb{R^3} \to \mathbb{R^3}$ such that $S(E)=B_1(0)$ and $E=S^{-1}(B_1(0))$.
So $\mathcal{H^s}(E)=|$det$(T^{-1})|\mathcal{H^s}(B_1(0))$
Here I don't see how to continue and I'm not sure if it works.
How to proceed to compute the Haussdorf measure and getting the dimension with the standard Basis?
Your triangle $T$ is a subset of the $xy$ plane. Its $2$ dimensional Hausdorff measure is just its area: $\dfrac 12$.
Likewise, the line segment $I_1$ lies in the $xz$ plane so its $1$ dimensional Hausdorff measure is its length: $\sqrt 2$. But $H^1(I_1) < \infty$ implies $H^2(I_1) = 0$. Likewise for $I_2$ and $I_3$. Then you get $$\frac 12 = H^2(T) \le H^2(E) \le H^2(I_1) + H^2(I_2) + H^2(I_3) + H^2(T) = H^2(T) = \frac 12$$ so that $H^2(E) = \dfrac 12$.