Suppose $X$ is a continuum (a compact connected Hausdorff space, not necessarily metrizable) of dimension one and $Y$ is a subcontinuum of $X$ (i.e. a subspace of $X$ which is a continuum). If the first Cech cohomology group $\check{H}^1(X;\mathbb{Z})$ is trivial, must it also be the case that $\check{H}^1(Y;\mathbb{Z})$ be trivial?
I don't believe this is true, but I can't seem to find a counterexample. The only dimension one continua with trivial first Cech cohomology group which I can think of at the moment are the pseudoarc and the arc, but both are hereditarily equivalent, so they won't work. I definitely could be completely wrong in my belief as I am very inexperienced in algebraic topology, so if there is a simple reason as to why this is true, then it would be great to know why.
In the answer I assume that by dimension you mean Lebesgue covering dimension (or the cohomological dimension).
There are no such continua. The reason is that for a finite-dimensional compact Hausdorff space, cohomological dimension equals covering dimension. On the other hand, integer cohomological dimension of a space $X$ is the least $n$ such that there exists a closed subset $A$ such that the pull-back map $$ p: \check{H}^n(X)\to \check{H}^n(A) $$
is surjective: see the discussion and references here).
A textbook proof of the equality of covering and cohomological dimensions (assuming that the space has finite covering dimension) can be found for instance in Theorem VIII.2 in
Nagata, J., Modern dimension theory. Rev. and ext. ed, Sigma Series in Pure Mathematics, Vol. 2. Berlin: Heldermann Verlag. IX, 284 p. (1983). ZBL0518.54002.
In your situation, if $\check{H}^1(X)=0$, then surjectivity of $p$ implies that for each closed subset $Y\subset X$ (regardless of its connectivity), $\check{H}^1(Y)=0$.