Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space, and (after an edit) for all infinite-dimensional Banach spaces.

There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$.

To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping: $$ f(a)_n = \frac{1}{n+1}a_n $$ this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$.

A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.

Added in edit:

As Bill Johnson points out, there is an injective bounded map from $\ell^2$ into any infinite-dimensional Banach space $E$. By the same argument used to transfer the example to $\ell^2$, this allows us to transfer the example to $E$.

In the other direction, the convex hull of a compact subset $K$ of a finite-dimensional space is compact (using Carathéodory's theorem we can express the convex hull of $K$ as the continuous image of the compact set $K^{d+1} \times P(d+1)$, where $d$ is the dimension. Therefore the $\sigma$-convex hull and closed convex hull of $K$ coincide.

All together, this means:

If $E$ is a Banach space, the statement "for all compact sets $K \subseteq E$, the closed convex hull equals the $\sigma$-convex hull" is equivalent to "$E$ is finite-dimensional".

There, however, complete locally convex spaces in which every bounded set, and therefore every compact set, is contained in a finite-dimensional subspace, and for which, therefore, the $\sigma$-convex and closed convex hulls of compact sets coincide. One example is the space $\phi$ of finitely supported functions $\mathbb{N} \rightarrow \mathbb{R}$, topologized as an $\mathbb{N}$-fold locally convex coproduct of $\mathbb{R}$ with itself, or equivalently as the strong dual space of $\mathbb{R}^{\mathbb{N}}$.

Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space, and (after an edit) for all infinite-dimensional Banach spaces.

There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$.

To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping: $$ f(a)_n = \frac{1}{n+1}a_n $$ this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$.

A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.

Added in edit:

As Bill Johnson points out, there is an injective bounded map from $\ell^2$ into any infinite-dimensional Banach space $E$. By the same argument used to transfer the example to $\ell^2$, this allows us to transfer the example to $E$.

In the other direction, the convex hull of a compact subset $K$ of a finite-dimensional space is compact (using Carathéodory's theorem we can express the convex hull of $K$ as the continuous image of the compact set $K^{d+1} \times P(d+1)$, where $d$ is the dimension. Therefore the $\sigma$-convex hull and closed convex hull of $K$ coincide.

All together, this means:

If $E$ is a Banach space, the statement "for all compact sets $K \subseteq E$, the closed convex hull equals the $\sigma$-convex hull" is equivalent to "$E$ is finite-dimensional".

There are, however, complete locally convex spaces in which every bounded set, and therefore every compact set, is contained in a finite-dimensional subspace, and for which, therefore, the $\sigma$-convex and closed convex hulls of compact sets coincide. One example is the space $\phi$ of finitely supported functions $\mathbb{N} \rightarrow \mathbb{R}$, topologized as an $\mathbb{N}$-fold locally convex coproduct of $\mathbb{R}$ with itself, or equivalently as the strong dual space of $\mathbb{R}^{\mathbb{N}}$.

Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space.

There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$.

To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping: $$ f(a)_n = \frac{1}{n+1}a_n $$ this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$.

A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.

Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space, and (after an edit) for all infinite-dimensional Banach spaces.

There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$.

To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping: $$ f(a)_n = \frac{1}{n+1}a_n $$ this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$.

A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.

Added in edit:

As Bill Johnson points out, there is an injective bounded map from $\ell^2$ into any infinite-dimensional Banach space $E$. By the same argument used to transfer the example to $\ell^2$, this allows us to transfer the example to $E$.

In the other direction, the convex hull of a compact subset $K$ of a finite-dimensional space is compact (using Carathéodory's theorem we can express the convex hull of $K$ as the continuous image of the compact set $K^{d+1} \times P(d+1)$, where $d$ is the dimension. Therefore the $\sigma$-convex hull and closed convex hull of $K$ coincide.

All together, this means:

If $E$ is a Banach space, the statement "for all compact sets $K \subseteq E$, the closed convex hull equals the $\sigma$-convex hull" is equivalent to "$E$ is finite-dimensional".

There, however, complete locally convex spaces in which every bounded set, and therefore every compact set, is contained in a finite-dimensional subspace, and for which, therefore, the $\sigma$-convex and closed convex hulls of compact sets coincide. One example is the space $\phi$ of finitely supported functions $\mathbb{N} \rightarrow \mathbb{R}$, topologized as an $\mathbb{N}$-fold locally convex coproduct of $\mathbb{R}$ with itself, or equivalently as the strong dual space of $\mathbb{R}^{\mathbb{N}}$.