One can start with the fundamental representation of $SU(5)$ as
\begin{equation}
\mathbf{[5]} = \begin{pmatrix} q_{i}\\ \ell_{j} \end{pmatrix} = \begin{pmatrix} q_{1}\\ q_{2}\\ q_{3}\\ \ell_{1} \\ \ell_{3} \end{pmatrix}
\end{equation}
where $q_{i}\,(i=1, 2, 3)$ and $\ell_{j}\,(j=1, 2)$ are the quark and lepton fields respectively.
Now, one can embed $SU(3)$ and $SU(2)$, in $SU(5)$ in the following way,
\begin{equation}
SU(5) = \begin{pmatrix} SU(3)_{c} & \\
& SU(2)_{L} \end{pmatrix}
\end{equation}
Then, following the transformation rule of fundamental representation $\mathbf{[5]}$, we can write
\begin{equation*}
\mathbf{[5]} \rightarrow \begin{pmatrix}
SU(3)_{c} & \\
& SU(2)_{L} \\
\end{pmatrix}
\begin{pmatrix}
q_{i} \\ \ell_{r}
\end{pmatrix}
\end{equation*}
Recall that quarks are triplet under $SU(3)_{c}$ and leptons are color blind, which simply put in the following way,
\begin{eqnarray}
\mathbf{[5]} & = & \mathbf{(3,1)}\oplus \mathbf{(1,2)} \\
\mathbf{[\bar{5}]} & = & \mathbf{(\bar{3},1)} \oplus \mathbf{(1,\bar{2})}
\end{eqnarray}
where $\mathbf{[\bar{5}]}$ is for anti-fundamental representation. Both up and down quark fields are triplets under the color group. One immediate question is, what fields corresponds to the $\mathbf{(\bar{3},1)}$ or $\mathbf{(3,1)}$?
Recall that one of the generators $SU(5)$ is electric charge $Q$ itself and which is diagonal! Therefore
\begin{equation}
\sum\mathrm{Eigenvalues} = 0
\end{equation}
which fixes $\mathbf{[\bar{3}]}$ in $\mathbf{[\bar{5}}]$ as down quarks.
\begin{equation*}
\bar{\mathbf{[5]}} = (\mathbf{3},\mathbf{1}) \oplus (\mathbf{1},\mathbf{2}) =
\begin{pmatrix}
d^{rc} & d^{bc} & d^{gc} & e^{-} & -\nu_e
\end{pmatrix}_L
\end{equation*}
From this one can immediately check
\begin{equation*}
3Q_{d^{c}}+Q_{e^{-}} = 0, \Longrightarrow Q_{d^{c}} = +\frac{1}{3}e
\end{equation*}
It says, fixing the charge of the electron automatically fixes the charge of the down-quarks. Which is indeed the quantization of electric charge. The remaining standard model matter fields can be contained in $[\mathbf{10}]$ representation. To achieve that we need to do some amount of group theory. Consider the following product
\begin{align}
\mathbf{[5]}\otimes\mathbf{[5]} &= [(\mathbf{3},\mathbf{1}_{2})\oplus (\mathbf{1}_{3},\mathbf{2})] \otimes [(\mathbf{3},\mathbf{1}_{2})\oplus (\mathbf{1}_{3},\mathbf{2})] \\
&= (\mathbf{3}\otimes\mathbf{3} ,\mathbf{1}_{2})\oplus (\mathbf{3},\mathbf{2})\oplus(\mathbf{3},\mathbf{2})\oplus (\mathbf{1_{3}},\mathbf{2}\otimes\mathbf{2})\\
&= (\mathbf{6},\mathbf{1}_{2})\oplus (\mathbf{\bar{3}},\mathbf{1}_{2})\oplus 2(\mathbf{3},\mathbf{2})\oplus (\mathbf{1_{3}},\mathbf{3})\oplus(\mathbf{1_{3}},\mathbf{1_{2}})
\end{align}
where we have used tensor product decomposition of $SU(3)$ and $SU(2)$ Lie algebra, namely
\begin{align}
\mathbf{[3]} \otimes \mathbf{[3]} &= \mathbf{[6]}_{S}\oplus\mathbf{[\bar{3}}_{A}]\\
\mathbf{[2]} \otimes \mathbf{[2]} &=
\mathbf{[3]}_{S}\oplus\mathbf{[1]}_{A}
\end{align}
Now, from the general rule of decomposition for $SU(n)$
\begin{equation}
\mathbf{[5]}\otimes\mathbf{[5]} = \mathbf{[10]_{A}}\oplus\mathbf{[15]_{S}}
\end{equation}
From there, one can immediately identify the anty-symmetric $\mathbf{[10]_{A}}$ as,
\begin{equation}
\mathbf{[10]_{A}} = (\mathbf{\bar{3}},\mathbf{1})\oplus (\mathbf{3},\mathbf{2})_{A}\oplus(\mathbf{1},\mathbf{3})_{A}
\end{equation}
From the representations of the SM field,
The $SU(3)_{c}\times SU(2)_{L}$ quantum numbers $(a,b)$ for fermionic states are:
\begin{align}
\begin{pmatrix} \nu_e & e^- \\ \end{pmatrix}_{L} &: (\mathbf{1},\mathbf{2}) \equiv (\mathbf{1},\bar{\mathbf{2}}) \\
e^{+}_{L} &: (\mathbf{1},\mathbf{1}) \\
\begin{pmatrix} u^{i} & d^{i} \\ \end{pmatrix}_{L} &: (\mathbf{3},\mathbf{2}) \\
u_L^{ic} &: (\bar{\mathbf{3}},\mathbf{1}) \\
d_L^{ic} &: (\bar{\mathbf{3}},\mathbf{1})
\end{align}
One can find how exactly they fit in $SU(5)$.
\begin{equation}
\mathrm{SM} = \mathbf{[\bar{5}]}\oplus\mathbf{[10]}
\end{equation}
Note. To see the (anti-)symmetric structure of the representations one has to play with Young tableaux.
