Let $f_i:X \to X_i$, $i \in I$ be a family of functions in topological spaces $X_i$, $A \subset X$ a subset and $\iota_A:A \to X$ the inclusion map. Suppose that $X$ has the initial topology with respect to the family $\{f_i\}_{i \in I}$.
I wanna show that the subspace topology of $A$ with the initial topology with respect to the family $\{f_i \circ \iota_A\}_{i \in I}$ the same is but I got stuck.
Some help would be really nice!
Let $\tau_A$ denote the subspace topology on $A$ and $\tau_F$ denote the initial topology given by $F=\{f_i \circ \iota_A: i \in I\}$. Since for all $i$ we have that $f_i \circ \iota_A \colon (A,\tau_A) \to X_i$ is continuous, (because composition of continuous gives continuous) then this shows that the initial topology $\tau_F$ is coarser than $\tau_A$ (i.e. $\tau_F \subseteq \tau_A$). Now take $U$ open in $\tau_A$ and $p \in U$. Then there exists an open set of $X$, say $O$ such that $U= A\cap O$. Since $p \in O$ there exists some open sets $U_{j_k} \subseteq X_{j_k}$, with $j_k \in I$ for all $k\in \{1,...,n\}$ such that $p \in f_{j_1}^{-1}[U_{j_1}] \cap \ldots \cap f_{j_n}^{-1}[U_{j_n}] \subseteq O$. Finally, note that $p \in (f_{j_1}\circ \iota_A)^{-1}[U_{j_1}] \cap \ldots \cap (f_{j_n}\circ\iota_A)^{-1}[U_{j_n}] \subseteq O\cap A$. This shows that $p$ is an interior point of $U$ with respect to the initial topology and hence $\tau_A \subseteq \tau_F$.
This is a consequence of the transitive law of initial topologies, that I showed in a general form here, to show that $\prod_i A_i$ as a subspace of the product has the same topology as the product topology of the subspace topologies.
I'll give a direct proof for your case:
The initial topology on $X$ wrt $\{f_i \mid i \in I\}$ has as its standard subbase
$$\mathcal{S} = \{f_i^{-1}[O]\mid i \in I, O \subseteq X_i \text{ open }\}$$
so the subspace topology on $A$ has subbase
$$\mathcal{S}_A= \{f_i^{-1}[O] \cap A\mid i \in I, O \subseteq X_i \text{ open }\}$$
Now note that $(f_i \circ \iota_A)^{-1}[O] = f_i^{-1}[O] \cap A$ as well, so that $\mathcal{S}_A$ is exactly the standard subbase for the initial topology on $A$ wrt the maps $\{f_i \circ \iota_A \mid i \in I\}$, as required.
The explicit description by subbases of these topologies is helpful here.
The other answers are good, but I personally prefer to prove things by using the universal property when possible, so here is another proof. There are two things to show:
The subspace topology on $A\subset X$ is the initial topology for the inclusion map $i_A:A\to X$.
In general, if you are given functions $f_i: Y\to Y_i$ and $f_{i,j}: Y_i\to Y_{i,j}$ where the $Y_{i,j}$ are topological spaces, then the initial topology on $Y$ defined by the family $(f_{i,j}\circ f_i: Y\to Y_{i,j})_{i,j}$ (topology $(i)$) is the same as the initial topology defined by $(f_i: Y\to Y_i)_i$ where each $Y_i$ has the initial topology defined by $(f_{i,j}: Y_i\to Y_{i,j})_j$ (topology $(ii)$); this is the transitivity property evoked by Henno Brandsma.
The first point can be taken as the definition of the subspace topology, but it is easy to see that it is equivalent to the more usual definition using open subsets (it is just saying that the subspace topology on $A$ is characterized by the fact that any map $B\to A$ is continuous if and only if it is continuous as a map $B\to X$).
For the second point, let $g:B\to Y$ be a function, where $B$ is a topological space. Then it is continuous for the topology $(i)$ iff each $f_{i,j}\circ f_i\circ g$ is continuous, by definition of the initial topology. And it is continuous for the topology $(ii)$ iff each $f_i\circ g$ is continuous, and by assumption on the topology of $Y_i$ this is equivalent to each $f_{i,j}\circ f_i\circ g$ being continuous. So the two conditions are the same, which means the two topologies coincide.
