Let $X=S^1$ and $\mathcal {F},\mathcal {G} $ two sheaves with
$\mathcal {F}(U)=${$f:U\rightarrow \mathbb {R} $} and
$\mathcal {G}(U)=${$f:U\rightarrow X$}, for any open set $U $ and
$\theta_U : \mathcal {F}(U)\rightarrow \mathcal {G}(U)$
$\theta_U(f)=\pi○f $
where $\pi: \mathbb {R}\rightarrow S^1$ is the projection.
I want to show that $U\rightarrow Im \theta_U $ is not a sheaf.
I'll take the projection $\pi : \mathbb{R} \to S^1$ to be the map defined by $x \mapsto e^{2\pi i x}$, so that all real numbers that differ by an integer lie over the same point on the circle.
Write $\mathcal{H}$ for the presheaf $U \mapsto \mathrm{im}(\theta_U)$.
Intuitively, can demonstrate that $\mathcal{H}$ is not a sheaf by piecing together functions which cause the circle to 'break' when you try to factor through the projection.
For example, let $U_x = S^1 \setminus \{ e^{2\pi i x} \}$ for all $x \in \mathbb{R}$. Then $\{ U_0, U_{\frac{1}{2}} \}$ is a cover of $S^1$. Define $f : U_0 \to \mathbb{R}$ by $f(e^{2\pi i t}) = t$ for all $0 < t < 1$, and define $g : U_{\frac{1}{2}} \to \mathbb{R}$ by $g(e^{2\pi i t}) = t$ for all $\frac{1}{2} < t < \frac{3}{2}$. Note that $\pi \circ f$ and $\pi \circ g$ agree on $U_0 \cap U_{\frac{1}{2}} = S^1 \setminus \{ 1, -1 \}$. The unique map $S^1 \to S^1$ extending $\pi \circ f$ and $\pi \circ g$ is $\mathrm{id}_{S^1}$, but this doesn't factor through $\pi: \mathbb{R} \to S^1$, so is not an element of $\mathcal{H}(S^1)$.