$X$ is compact Hausdorff connected 1-dimensional topology space.
Dimension here refers to covering dimension.
$M_n(C(X))\simeq C(X:M_n)$ is the space of continuous maps from $X$ to $M_n(\mathbb C)$, the $n\times n$ complex matrix algebra.
Say a projection in $M_n(C(X))$ is trivial if $p(t)=p(0)$ for all $t\in X$.
I think this is true when $X$ is 1-dimensional finite CW complex. Being a 1-dimensional finite CW complex means it is formed with finitely many points and curve segments. And, for every two curve segments, their intersection is a finite(or empty) set.
Here is my brief explanation.
1, Two projections that satisfy $\|p-q\|<1$ are unitarily equivalent.
2, Projection $f\in M_n(C([0,1]))$ is unitarily equivalent to some trivial projection, since it can be cut into finitely many intervals on each of which the variation of $f$ is less than $1$.
3,All unitaries in $M_m(\mathbb C)$ are homotopy.
4,Suppose further that $u^*f(0)u=v^*f(1)v$, then the $g$ that makes $g^*fg$ trivial can be chosen to satisfy $g(0)=u$ and $g(1)=v$.
5,Induct on the number of curve segments. Assume $X$ is an 1-dimensional finite CW complex and $f\in M_n(C(X))$ is unitarily equivalent to some trivial projection. Let $g^*fg$ be trivial. Suppose now a curve segment $L\simeq [0,1]$ is attached to $X$ and $f'$ is an extension of $f$ to $X\cup L$. Then there is a $g'$ on $L$ such that $g'^*fg'$ is trivial and $g'(L_0)=g(L_0)$ and $g'(L_1)=g(L_1)$. So $g\cup g'$ is the desired unitary.
I have also been told this is not true for $\text{dim}(X)\geq 2$ in general. So I wonder if it will be true under the condition that $\text{dim}(X)=1$.
I wanted to show that any unital hereditary subalgebra of $C(X)\otimes F$, where $X$ is an 1-dimensional finite CW complex and $F$ is a finite dimensional C*-algebra, is again of form $C(X)\otimes F'$.
A unital hereditary subalgebra can be written as $p(C(X)\otimes F)p$. Since $F$ is finite, it is $p_1(M_{n_1}(C(X)))p_1\oplus ...\oplus p_k(M_{n_k}(C(X)))p_k $. Then using the above proposition, it is isomorphic to $M_{m_1}(C(X))\oplus ...\oplus M_{m_k}(C(X))\simeq C(X)\otimes F'$.