It seems to me that that this is true:
In the category of Sets, transfinite composition of monomorphisms is again a monomorphism.
Explicitly, given a $\lambda$-sequence $$X_0 \xrightarrow{f_1} X_1 \cdots $$ Each map $f_i$ is a monomorphism, then is the canonical $$ g:X_0 \rightarrow colim_{i < \lambda} X_i$$ A monomorphism?
Is this true more generally in other categories?
My thoughts: In sets there are explicit description of colimit. So I suppose we do this by some kind of ordinal induction?
We know this is true for successor ordinals. $\beta \le \lambda$. If $\beta\le \lambda$ is a limit ordinal, then we have $$g:X_0 \rightarrow X_\beta \simeq colim_{j< \beta} X_j$$ By property of ordinals, being totally ordered, exists some $k <\beta$, such that $$f_k(x)=f_k(y) \in X_k$$ By hypothesis, $x=y$.
Yes, this is true for sets, and your proof sketch is basically correct. By induction, you show the transfinite composition $X_0\to X_i$ is injective for all $i$. Successor steps are trivial (you just compose with the injection $X_i\to X_{i+1}$). At limit steps, you use the fact that if two elements of $X_i=\operatorname{colim}_{j<i} X_j$ are equal, they must be represented by equal elements of $X_j$ for some $j<i$. So if two elements of $X_0$ map to the same element of $X_i$, they must have mapped to the same element of $X_j$ for some $j<i$, but by the inductive hypothesis the map $X_0\to X_j$ is injective.
It is not true in arbitrary categories. For instance, in the category of finite abelian groups, let $X_n=\mathbb{Z}/(p^n)$ for some $p>1$ with maps $X_n\to X_{n+1}$ given by multiplication by $p$. All these maps are injections, but the colimit $X_\omega$ is trivial, since there are no nontrivial maps from the diagram to any finite abelian group (an element in the image of such a map would have to be annihilated by some power of $p$ but also infinitely divisible by $p$, which is not possible for a nonzero element in a finite abelian group). So in particular, the transfinite composition $X_1\to X_\omega$ is not monic.