[proofplan]
We first prove directly that the image of the connected set $C$ under the continuous map $f$ is connected, avoiding any external citation. Then we construct the connected component of $Y$ containing one point of $f(C)$ as the union of all connected subsets of $Y$ containing that point. Since this union is connected and maximal among connected subsets of $Y$, it is a connected component, and it contains $f(C)$.
[/proofplan]
[step:Show that the image of $C$ under $f$ is connected]
Because $C$ is a connected component of $X$, the subset $C \subset X$ is connected and nonempty. Define the restricted map
\begin{align*}
g: C &\to f(C) \\
x &\mapsto f(x),
\end{align*}
where $C$ carries the [subspace topology](/page/Subspace%20Topology) inherited from $X$ and $f(C)$ carries the subspace topology inherited from $Y$.
The map $g$ is continuous: if $O \subset f(C)$ is open in the subspace topology, then there exists $V \in \tau_Y$ such that $O = V \cap f(C)$, and
\begin{align*}
g^{-1}(O) = C \cap f^{-1}(V),
\end{align*}
which is open in $C$ because $f$ is continuous.
Suppose, for contradiction, that $f(C)$ is disconnected. Then there exist nonempty subsets $A, B \subset f(C)$ that are open in the subspace topology on $f(C)$, disjoint, and satisfy
\begin{align*}
f(C) = A \cup B.
\end{align*}
Since $g$ is continuous, the subsets $g^{-1}(A)$ and $g^{-1}(B)$ are open in $C$. They are disjoint and satisfy
\begin{align*}
C = g^{-1}(A) \cup g^{-1}(B).
\end{align*}
They are also nonempty because $A$ and $B$ are nonempty subsets of $f(C)$ and $g: C \to f(C)$ is surjective by definition of $f(C)$. Thus $C$ is separated into two nonempty disjoint open subsets, contradicting the connectedness of $C$. Therefore $f(C)$ is connected.
[guided]
Because $C$ is a connected component of $X$, it is a maximal connected subset of $X$; in particular, it is connected and nonempty. We want to prove that $f(C)$ is connected. The natural map to study is the restriction of $f$ to $C$, but with codomain reduced to its actual image. Define
\begin{align*}
g: C &\to f(C) \\
x &\mapsto f(x),
\end{align*}
where $C$ has the subspace topology from $X$ and $f(C)$ has the subspace topology from $Y$.
First we verify continuity of $g$. Let $O \subset f(C)$ be open in the subspace topology on $f(C)$. By the definition of the subspace topology, there exists an [open set](/page/Open%20Set) $V \in \tau_Y$ such that
\begin{align*}
O = V \cap f(C).
\end{align*}
Then the preimage of $O$ under $g$ is
\begin{align*}
g^{-1}(O) = C \cap f^{-1}(V).
\end{align*}
Since $f: X \to Y$ is continuous and $V$ is open in $Y$, the set $f^{-1}(V)$ is open in $X$. Therefore $C \cap f^{-1}(V)$ is open in the subspace topology on $C$. Hence $g$ is continuous.
Now suppose, toward a contradiction, that $f(C)$ is disconnected. Then there are subsets $A, B \subset f(C)$ such that $A$ and $B$ are nonempty, open in the subspace topology on $f(C)$, disjoint, and cover $f(C)$:
\begin{align*}
f(C) = A \cup B.
\end{align*}
Taking preimages under the continuous map $g$, the subsets $g^{-1}(A)$ and $g^{-1}(B)$ are open in $C$. They are disjoint because $A$ and $B$ are disjoint, and they cover $C$ because $A \cup B = f(C)$:
\begin{align*}
C = g^{-1}(A) \cup g^{-1}(B).
\end{align*}
They are nonempty because $A$ and $B$ are nonempty subsets of $f(C)$, and every point of $f(C)$ is equal to $g(x)$ for some $x \in C$. Thus $C$ has been written as a union of two nonempty disjoint open subsets, which contradicts the connectedness of $C$. Therefore $f(C)$ is connected.
[/guided]
[/step]
[step:Construct the connected component of $Y$ containing a point of $f(C)$]
Choose $x_0 \in C$, and define $y_0 := f(x_0) \in f(C)$. Let $\mathcal{S}$ denote the collection of all connected subsets of $Y$ that contain $y_0$:
\begin{align*}
\mathcal{S} := \{E \subset Y : E \text{ is connected and } y_0 \in E\}.
\end{align*}
Define
\begin{align*}
D := \bigcup_{E \in \mathcal{S}} E.
\end{align*}
We show that $D$ is connected. Suppose, for contradiction, that $D$ is disconnected. Then there exist nonempty disjoint subsets $A, B \subset D$, open in the subspace topology on $D$, such that
\begin{align*}
D = A \cup B.
\end{align*}
Since $y_0 \in D$, either $y_0 \in A$ or $y_0 \in B$. Assume $y_0 \in A$; the other case is identical with $A$ and $B$ interchanged.
For each $E \in \mathcal{S}$, the intersections $E \cap A$ and $E \cap B$ are open in the subspace topology on $E$, disjoint, and cover $E$. Since $E$ is connected and $y_0 \in E \cap A$, we must have $E \cap B = \varnothing$. Therefore every $E \in \mathcal{S}$ is contained in $A$, and hence
\begin{align*}
D = \bigcup_{E \in \mathcal{S}} E \subset A.
\end{align*}
This contradicts the assumption that $B$ is nonempty. Hence $D$ is connected.
By construction, $D$ is a connected subset of $Y$ containing $y_0$. If $D' \subset Y$ is connected and $D \subset D'$, then $y_0 \in D'$, so $D' \in \mathcal{S}$ and therefore $D' \subset D$. Hence $D' = D$. Thus $D$ is maximal among connected subsets of $Y$, so $D$ is a connected component of $Y$.
[/step]
[step:Place $f(C)$ inside the constructed component]
From the first step, $f(C)$ is connected. Since $y_0 = f(x_0)$ and $x_0 \in C$, we have $y_0 \in f(C)$. Therefore $f(C) \in \mathcal{S}$. By the definition of $D$ as the union of all sets in $\mathcal{S}$,
\begin{align*}
f(C) \subset D.
\end{align*}
The subset $D \subset Y$ is a connected component of $Y$, so $f(C)$ is contained in a connected component of $Y$.
[/step]
