You are not explicit whether "compact" includes Hausdorff, so it is not absolutely clear what the compact-open topology is. Let us ignore this point as it is irrelevant in the sequel.
Let us first understand the exponential map on the level of sets. For sets $A, B$ let $[A,B]$ denote the set of all functions $f : A \to B$. Define $e = e^{A,B,C}: [A \times B,C] \to [A,[B,C]]$ as follows: For $f : A \times B \to C$ and $a \in A$ define a function $f_a : B \to C, f_a(b) = f(a,b)$. Then $e(f)$ is defined by $e(f)(a) = f_a$. It is easy to verify that $e$ is a bijection of sets.
Now we come to topology. Given topological spaces $X, Y, Z$, it is clear that if $f : X \times Y \to Z$ is continuous, then all $f_x : Y \to Z$ are continuous, i.e. $f_x \in \text{Hom}(Y,Z)$. If we endow $\text{Hom}(Y,Z)$ with the compact-open topology, we get the space $Z^Y = \text{Maps}(Y,Z)$. It is easy to verify that $e(f) \in \text{Hom}(X,Z^Y)$. Thus we obtain an injective function
$$e : \text{Hom}(X \times Y,Z) \to \text{Hom}(X,Z^Y) .$$
It is well-known that if $Y$ is locally compact, then $e$ is bijective for all $X,Z$. At this point it becomes relevant how compact and locally compact are defined. However, for a Hausdorff $Y$ all definitions yield the same result.
A slightly more general framework assuring that $e$ is bijective for all $X,Z$ is this. The evaluation map is defined by $\Omega : Z^Y \times Y \to Z, \Omega(f,y) = f(y)$. The following are well-known:
Given $Y,Z$, $e$ is bijective for all $X$ iff $\Omega$ is continuous.
If $Y$ is locally compact, then $\Omega$ is continuous for all $Z$.
What happens if we have $Y,Z$ such that $\Omega$ is not continuous? Then for $X = Z^Y$ the map $e$ is not bijective. To see this, consider the identity map $id \in \text{Hom}(Z^Y,Z^Y)$. If $e$ were bijective, then $e(\phi) = id$ for some $\phi \in \text{Hom}(Z^Y \times Y,Z)$. By definition of $e$ we must have $\phi = \Omega$ which is a contradiction.
Thus, to show that for a space $Y$ the map $e$ is not bijective for all $X,Z$, it suffices to find $Z$ such that $\Omega$ is not continuous and take $X = Y^Z$. Clearly,
such $Y$ cannot be locally compact.
Your question says that $Y = \mathbb{R} \setminus \{ \frac{1}{n} \mid n \in \mathbb{N}\}$ will do. Let us show that $\Omega$ is not continuous for $Z = I = [0,1]$.
Let $c \in I^Y$ be the constant function $c(y) = 0$. We claim that $\Omega$ is not continuous in $(c,0)$. We do it by contradiction. If $\Omega$ were continuous in $(c,0)$, then we can find $\varepsilon > 0$, compact $K_i \subset Y$ and open $U_i \subset I$ such that $c \in \langle K_i,U_i \rangle = \{ f \in I^Y \mid (f(K_i) \subset U_i \}$ and $\Omega(\bigcap_{i=1}^m \langle K_i,U_i \rangle \times U_\varepsilon) \subset [0,1)$. Here $U_\varepsilon = \{ y \in Y \mid \lvert y \rvert < \varepsilon \}$. Note that necessarily $0 \in U_i$ since $0 \in c(K_i)$. $K = \bigcup_{i=1}^m K_i$ is compact and $U = \bigcap_{i=1}^m U_i$ an open neighborhood of $0$, thus $c \in \langle K,U \rangle \subset \bigcap_{i=1}^m \langle K_i,U_i \rangle$ and $\Omega(\langle K,U \rangle \times U_\varepsilon) \subset [0,1)$.
Let $n \in \mathbb N$ such that $\frac{1}{n} < \varepsilon$. Since $(\frac{1}{n+1},\frac{1}{n})$ is closed in $Y$, the set $C = (\frac{1}{n+1},\frac{1}{n}) \cap K$ is compact, thus we find $\xi \in (\frac{1}{n+1},\frac{1}{n}) \setminus C$. Let $g : (\frac{1}{n+1},\frac{1}{n}) \to I$ be a continuous map such that $g(\xi)= 1$ and $g(x) = 0$ for $x \in C$. Extend $g$ to a continuous $f : Y \to I$ via $f(x) = 0$ for $x \notin (\frac{1}{n+1},\frac{1}{n})$. Then $f(K) = \{0\} \subset U$, i.e. $f \in \langle K,U \rangle$. We have $\xi \in U_\varepsilon$, thus $(f,\xi) \in \langle K,U \rangle \times U_\varepsilon$ and $\Omega(f,\xi) = f(\xi) = 1$. This contradicts $\Omega(\langle K,U \rangle \times U_\varepsilon) \subset [0,1)$.
Remark 1: We can generalize this as follows. Let $Y$ be a Tychonoff space (= completely regular Hausdorff space) which is not locally compact. Then the evaluation map $\Omega : I^Y \times Y \to I$ is not continuous.
To prove this, let $y_0 \in Y$ be a point which does not have a compact neighborhood. Let $c \in I^Y$ be the constant function $c(y) = 0$. We claim that $\Omega$ is not continuous in $(c,y_0)$. We do it by contradiction. If $\Omega$ were continuous in $(c,y_0)$, then we can find an open neighborhood $V$ of $y_0$ in $Y$, compact $K_i \subset Y$ and open $U_i \subset I$ such that $c \in \langle K_i,U_i \rangle$ and $\Omega(\bigcap_{i=1}^m \langle K_i,U_i \rangle \times V) \subset [0,1)$. Note that necessarily $0 \in U_i$ since $0 \in c(K_i)$. $K = \bigcup_{i=1}^m K_i$ is compact and $U = \bigcap_{i=1}^m U_i$ an open neighborhood of $0$, thus $c \in \langle K,U \rangle \subset \bigcap_{i=1}^m \langle K_i,U_i \rangle$ and $\Omega(\langle K,U \rangle \times V) \subset [0,1)$.
We have $V \setminus K \ne \emptyset$, because otherwise $V \subset K$ so that $K$ would be a compact neighborhood of $y_0$. Pick $\eta \in V \setminus K$ and let $f: Y \to I$ be a continuous map such that $f(y) = 0$ for $y \in K$ and $f(\eta) = 1$. Then $f(K) = \{0\} \subset U$, i.e. $f \in \langle K,U \rangle$. We have $\eta \in V$, thus $(f,\eta) \in \langle K,U \rangle \times V$ and $\Omega(f,\eta) = f(\eta) = 1$. This contradicts $\Omega(\langle K,U \rangle \times V) \subset [0,1)$.
Remark 2: If we take $S$= Sierpinski space, then for any Hausdorff space $Y$ which is not locally compact the evaluation map $\Omega : S^Y \times Y \to S$ is not continuous. The proof is similar as above.
