I'm struggling to show that in a type II$_1$ and any $n$ $\exists$ a subfactor $M$ such that $M \cong M_n$ .
I suppose it should follow from the isomorphism between the equivalence classes of projections and the interval $[0,\operatorname{tr}(1)]$ but i can not figure out the details
Say your II$_1$-factor is $N$. You can always find projections $p_1,\ldots,p_n$ with $\sum_jp_j=I $ and $\tau(p_j)=1/n$. Because you are in a factor and they have equal trace, these projections are pairwise equivalent. In particular there exist partial isometries $v_1,\ldots,v_n$ such that $v_j^*v_j=p_j$, $v_jv_j^*=p_1$.
Next you define $$ e_{kj}=v_k^*v_j,\ \ \ k,j=1,\ldots,n. $$ You have $$\tag1 e_{kj}e_{st}=v_k^*v_jv_s^*v_t=v_k^*v_jp_jp_sv_s^*v_t=\delta_{j,s}\,v_k^*v_jv_j^*v_t^* =\delta_{j,s}\,v_k^*p_1v_t=\delta_{j,s}\,v_k^*v_t=\delta_{j,s}\,e_{kt}. $$ Now let $M=\operatorname{span}\{e_{kj},\ k,j=1,\ldots,n\}\subset N$. Then the map $\pi:M\to M_n(\mathbb C)$ given by $\pi(e_{kj})=E_{kj}$ is a $*$-isomorphism. This is easily checked using $(1)$.